[proofplan] We prove the estimate directly from the definition of Hausdorff distance. If $K$ is close to $K'$ and $L$ is close to $L'$, then every point $k+l \in K+L$ can be approximated by a point $k'+l' \in K'+L'$ with the two approximation errors adding. Repeating the same argument with primed and unprimed sets exchanged gives the reverse containment estimate, and the Hausdorff bound follows. Continuity is then the sequential consequence of this Lipschitz estimate. [/proofplan]

[step:Verify that Minkowski sums of convex bodies remain convex bodies] We first check that all Hausdorff distances appearing below are distances between nonempty compact sets, and that Minkowski addition is a map into the space of convex bodies. Let \begin{align*} \alpha: \mathbb{R}^n \times \mathbb{R}^n &\to \mathbb{R}^n \\ (u,v) &\mapsto u+v \end{align*} denote the Euclidean addition map. Since $K \times L$ and $K' \times L'$ are nonempty compact subsets of $\mathbb{R}^n \times \mathbb{R}^n$ and $\alpha$ is continuous, the sets \begin{align*} K+L = \alpha(K \times L), \qquad K'+L' = \alpha(K' \times L') \end{align*} are nonempty compact subsets of $\mathbb{R}^n$. The set $K+L$ is convex: if $x_0,x_1 \in K+L$ and $t \in [0,1]$, choose $k_0,k_1 \in K$ and $l_0,l_1 \in L$ such that $x_i=k_i+l_i$ for $i \in \{0,1\}$. Then \begin{align*} (1-t)x_0+tx_1 = ((1-t)k_0+tk_1)+((1-t)l_0+tl_1) \in K+L, \end{align*} because $K$ and $L$ are convex. To prove convexity of $K'+L'$, let $y_0,y_1 \in K'+L'$ and $t \in [0,1]$. Choose $k'_0,k'_1 \in K'$ and $l'_0,l'_1 \in L'$ such that $y_i=k'_i+l'_i$ for $i \in \{0,1\}$. Then \begin{align*} (1-t)y_0+ty_1 =((1-t)k'_0+tk'_1)+((1-t)l'_0+tl'_1)\in K'+L', \end{align*} because $K'$ and $L'$ are convex. Thus $K'+L'$ is convex. Finally, $K+L$ has nonempty interior. For a set $E \subset \mathbb R^n$, let $\operatorname{int}(E)$ denote its topological interior in $\mathbb R^n$, and for $c \in \mathbb R^n$ and $\rho>0$ define the open Euclidean ball \begin{align*} B(c,\rho):=\{z\in\mathbb R^n: |z-c|<\rho\}. \end{align*} Choose $a \in \operatorname{int}(K)$ and $b \in \operatorname{int}(L)$. There exist $\rho>0$ and $\sigma>0$ such that $B(a,\rho) \subset K$ and $B(b,\sigma) \subset L$. If $\tau := \min\{\rho,\sigma\}$ and $z \in B(a+b,\tau)$, then $z=a+b+h$ for some $h \in \mathbb{R}^n$ with $|h|<\tau$, hence $a+h \in K$ and $b \in L$, so $z=(a+h)+b \in K+L$. Thus $B(a+b,\tau) \subset K+L$. Choose $a' \in \operatorname{int}(K')$ and $b' \in \operatorname{int}(L')$. There exist $\rho'>0$ and $\sigma'>0$ such that $B(a',\rho') \subset K'$ and $B(b',\sigma') \subset L'$. If $\tau' := \min\{\rho',\sigma'\}$ and $z' \in B(a'+b',\tau')$, then $z'=a'+b'+h'$ for some $h' \in \mathbb{R}^n$ with $|h'|<\tau'$, hence $a'+h' \in K'$ and $b' \in L'$, so $z'=(a'+h')+b' \in K'+L'$. Thus $B(a'+b',\tau') \subset K'+L'$. Therefore $K+L$ and $K'+L'$ are convex bodies. [/step]

[step:Define the two Hausdorff errors and approximate one summand at a time]Set \begin{align*} r := d_H(K,K'), \qquad s := d_H(L,L'). \end{align*} Since $K,K',L,L'$ are compact nonempty subsets of $\mathbb{R}^n$, both $r$ and $s$ are finite. The Lipschitz estimate below uses only this nonempty compactness; convexity and nonempty interior were needed in the preceding step to keep the operation inside $\mathcal{K}^n$. For each nonempty set $E \subset \mathbb{R}^n$, define the distance-to-$E$ function \begin{align*} \operatorname{dist}(\cdot,E): \mathbb{R}^n &\to [0,\infty) \\ z &\mapsto \inf_{w \in E} |z-w|. \end{align*} Let $x \in K+L$. By the definition of Minkowski sum, there exist $k \in K$ and $l \in L$ such that \begin{align*} x = k+l. \end{align*} Let $\varepsilon > 0$. Since $d_H(K,K') = r$, the definition of Hausdorff distance gives a point $k'_\varepsilon \in K'$ satisfying \begin{align*} |k-k'_\varepsilon| \leq r+\varepsilon. \end{align*} Similarly, since $d_H(L,L') = s$, there exists $l'_\varepsilon \in L'$ satisfying \begin{align*} |l-l'_\varepsilon| \leq s+\varepsilon. \end{align*} Define \begin{align*} x'_\varepsilon := k'_\varepsilon + l'_\varepsilon. \end{align*} Then $x'_\varepsilon \in K'+L'$, and the triangle inequality gives \begin{align*} |x-x'_\varepsilon| &= |(k+l)-(k'_\varepsilon+l'_\varepsilon)| \\ &= |(k-k'_\varepsilon)+(l-l'_\varepsilon)| \\ &\leq |k-k'_\varepsilon|+|l-l'_\varepsilon| \\ &\leq r+s+2\varepsilon. \end{align*} Therefore \begin{align*} \operatorname{dist}(x,K'+L') \leq r+s+2\varepsilon. \end{align*} Since $\varepsilon>0$ was arbitrary, \begin{align*} \operatorname{dist}(x,K'+L') \leq r+s. \end{align*}[/step]

[guided]The goal is to compare a typical point of $K+L$ with some point of $K'+L'$. Choose $x \in K+L$. By definition of the Minkowski sum, $x$ has a decomposition \begin{align*} x = k+l \end{align*} for some $k \in K$ and $l \in L$. Now define the two Hausdorff errors \begin{align*} r := d_H(K,K'), \qquad s := d_H(L,L'). \end{align*} For each nonempty set $E \subset \mathbb{R}^n$, we use the distance-to-$E$ function \begin{align*} \operatorname{dist}(\cdot,E): \mathbb{R}^n &\to [0,\infty) \\ z &\mapsto \inf_{w \in E} |z-w|. \end{align*} This definition says that $\operatorname{dist}(z,E)$ is the smallest distance from $z$ to points of $E$, understood as an infimum. The meaning of $r$ is that every point of $K$ lies within distance at most $r$ of $K'$, up to an arbitrarily small error if we do not want to discuss attainment of the closest point. Thus, for every $\varepsilon>0$, there exists $k'_\varepsilon \in K'$ such that \begin{align*} |k-k'_\varepsilon| \leq r+\varepsilon. \end{align*} Likewise, there exists $l'_\varepsilon \in L'$ such that \begin{align*} |l-l'_\varepsilon| \leq s+\varepsilon. \end{align*} We combine these two approximations in the only natural way: define \begin{align*} x'_\varepsilon := k'_\varepsilon + l'_\varepsilon. \end{align*} Since $k'_\varepsilon \in K'$ and $l'_\varepsilon \in L'$, the point $x'_\varepsilon$ belongs to $K'+L'$. The distance from $x$ to this approximating point is controlled by the triangle inequality: \begin{align*} |x-x'_\varepsilon| &= |(k+l)-(k'_\varepsilon+l'_\varepsilon)| \\ &= |(k-k'_\varepsilon)+(l-l'_\varepsilon)| \\ &\leq |k-k'_\varepsilon|+|l-l'_\varepsilon| \\ &\leq r+s+2\varepsilon. \end{align*} Because $\operatorname{dist}(x,K'+L')$ is the infimum of $|x-y|$ over all $y \in K'+L'$, this gives \begin{align*} \operatorname{dist}(x,K'+L') \leq r+s+2\varepsilon. \end{align*} Letting $\varepsilon \downarrow 0$ yields \begin{align*} \operatorname{dist}(x,K'+L') \leq r+s. \end{align*}[/guided]

[step:Take the supremum over $K+L$] The preceding step holds for every $x \in K+L$. Therefore \begin{align*} \sup_{x \in K+L} \operatorname{dist}(x,K'+L') \leq r+s. \end{align*} [/step]

[step:Repeat the approximation in the reverse direction]Let $y \in K'+L'$. Then there exist $k' \in K'$ and $l' \in L'$ such that \begin{align*} y = k'+l'. \end{align*} For every $\varepsilon>0$, the definition of $r=d_H(K,K')$ gives $k_\varepsilon \in K$ with \begin{align*} |k'-k_\varepsilon| \leq r+\varepsilon, \end{align*} and the definition of $s=d_H(L,L')$ gives $l_\varepsilon \in L$ with \begin{align*} |l'-l_\varepsilon| \leq s+\varepsilon. \end{align*} Set \begin{align*} y_\varepsilon := k_\varepsilon+l_\varepsilon. \end{align*} Then $y_\varepsilon \in K+L$, and the triangle inequality gives \begin{align*} |y-y_\varepsilon| &= |(k'+l')-(k_\varepsilon+l_\varepsilon)| \\ &\leq |k'-k_\varepsilon|+|l'-l_\varepsilon| \\ &\leq r+s+2\varepsilon. \end{align*} Letting $\varepsilon \downarrow 0$ gives \begin{align*} \operatorname{dist}(y,K+L) \leq r+s. \end{align*} Taking the supremum over $y \in K'+L'$ yields \begin{align*} \sup_{y \in K'+L'} \operatorname{dist}(y,K+L) \leq r+s. \end{align*}[/step]

[guided]We now prove the opposite one-sided estimate. Choose an arbitrary point $y \in K'+L'$. By the definition of Minkowski sum, there are points $k' \in K'$ and $l' \in L'$ such that \begin{align*} y = k'+l'. \end{align*} The Hausdorff error $r=d_H(K,K')$ controls points of $K'$ by points of $K$ as well as points of $K$ by points of $K'$, because Hausdorff distance is the maximum of the two directed distances. Therefore, for every $\varepsilon>0$, there exists $k_\varepsilon \in K$ with \begin{align*} |k'-k_\varepsilon| \leq r+\varepsilon. \end{align*} Similarly, since $s=d_H(L,L')$, there exists $l_\varepsilon \in L$ with \begin{align*} |l'-l_\varepsilon| \leq s+\varepsilon. \end{align*} Define \begin{align*} y_\varepsilon := k_\varepsilon+l_\varepsilon. \end{align*} Then $y_\varepsilon \in K+L$. Applying the triangle inequality in $\mathbb{R}^n$ gives \begin{align*} |y-y_\varepsilon| &= |(k'+l')-(k_\varepsilon+l_\varepsilon)| \\ &\leq |k'-k_\varepsilon|+|l'-l_\varepsilon| \\ &\leq r+s+2\varepsilon. \end{align*} Because $y_\varepsilon$ is one admissible point in $K+L$, the distance from $y$ to $K+L$ satisfies \begin{align*} \operatorname{dist}(y,K+L) \leq r+s+2\varepsilon. \end{align*} Letting $\varepsilon \downarrow 0$ gives \begin{align*} \operatorname{dist}(y,K+L) \leq r+s. \end{align*} Since this holds for every $y \in K'+L'$, taking the supremum gives \begin{align*} \sup_{y \in K'+L'} \operatorname{dist}(y,K+L) \leq r+s. \end{align*}[/guided]

[step:Combine the two one-sided bounds to obtain the Hausdorff estimate]By the definition of the Hausdorff distance between nonempty compact subsets of $\mathbb{R}^n$, \begin{align*} d_H(K+L,K'+L') = \max\left\{ \sup_{x \in K+L} \operatorname{dist}(x,K'+L'), \sup_{y \in K'+L'} \operatorname{dist}(y,K+L) \right\}. \end{align*} The two previous steps bound both entries in this maximum by $r+s$. Hence \begin{align*} d_H(K+L,K'+L') \leq r+s. \end{align*} Substituting the definitions of $r$ and $s$ gives \begin{align*} d_H(K+L,K'+L') \leq d_H(K,K') + d_H(L,L'). \end{align*}[/step]

[guided]The first step proved that $K+L$ and $K'+L'$ are nonempty compact subsets of $\mathbb{R}^n$, so the Hausdorff distance between them is computed by the standard formula \begin{align*} d_H(K+L,K'+L') = \max\left\{ \sup_{x \in K+L} \operatorname{dist}(x,K'+L'), \sup_{y \in K'+L'} \operatorname{dist}(y,K+L) \right\}. \end{align*} The first directed supremum is at most $r+s$ by the estimate for points of $K+L$. The second directed supremum is at most $r+s$ by the reverse estimate for points of $K'+L'$. Hence the maximum of the two quantities is also at most $r+s$: \begin{align*} d_H(K+L,K'+L') \leq r+s. \end{align*} Finally, $r$ and $s$ were defined by \begin{align*} r := d_H(K,K'), \qquad s := d_H(L,L'), \end{align*} so substituting these definitions gives \begin{align*} d_H(K+L,K'+L') \leq d_H(K,K') + d_H(L,L'). \end{align*}[/guided]

[step:Deduce continuity of Minkowski addition]Let $(K_m)_{m=1}^{\infty}$ and $(L_m)_{m=1}^{\infty}$ be sequences in $\mathcal{K}^n$ such that \begin{align*} d_H(K_m,K) \to 0, \qquad d_H(L_m,L) \to 0. \end{align*} Applying the estimate already proved with $K'=K_m$ and $L'=L_m$ gives \begin{align*} d_H(K_m+L_m,K+L) \leq d_H(K_m,K)+d_H(L_m,L). \end{align*} The right-hand side tends to $0$, so \begin{align*} d_H(K_m+L_m,K+L) \to 0. \end{align*} The first step shows that $A(K,L)=K+L$ is a convex body whenever $K$ and $L$ are convex bodies, so the map \begin{align*} A: \mathcal{K}^n \times \mathcal{K}^n &\to \mathcal{K}^n \\ (K,L) &\mapsto K+L \end{align*} is well-defined. The convergence just proved shows that $A$ is continuous with respect to the Hausdorff metric. This completes the proof.[/step]

[guided]To prove continuity, take sequences $(K_m)_{m=1}^{\infty}$ and $(L_m)_{m=1}^{\infty}$ in $\mathcal{K}^n$ with \begin{align*} d_H(K_m,K) \to 0, \qquad d_H(L_m,L) \to 0. \end{align*} The Lipschitz estimate already proved applies to the pair $(K_m,L_m)$ and the pair $(K,L)$, so \begin{align*} d_H(K_m+L_m,K+L) \leq d_H(K_m,K)+d_H(L_m,L). \end{align*} The right-hand side tends to $0$ because it is the sum of two real sequences tending to $0$. Therefore \begin{align*} d_H(K_m+L_m,K+L) \to 0. \end{align*} The first step also proved the well-definedness of the operation on convex bodies: if $K,L \in \mathcal{K}^n$, then $K+L$ is again nonempty, compact, convex, and has nonempty interior. Hence \begin{align*} A: \mathcal{K}^n \times \mathcal{K}^n &\to \mathcal{K}^n \\ (K,L) &\mapsto K+L \end{align*} is a well-defined map, and the sequential estimate proves that it is continuous for the Hausdorff metric.[/guided]