[proofplan] We prove the path-connectedness criterion directly. Given two points $x,y \in C$, we join them by the straight-line segment $\gamma(t)=(1-t)x+ty$. Convexity places the whole segment inside $C$, and an explicit Lipschitz estimate proves continuity into $\mathbb R^n$; the [subspace topology](/page/Subspace%20Topology) then upgrades this to continuity as a map into $C$. [/proofplan] [step:Choose arbitrary endpoints and define the line-segment map] If $C=\varnothing$, then there are no pairs $x,y \in C$, so the asserted pairwise path condition is vacuous. Now let $x,y \in C$ be arbitrary. Define a map \begin{align*} \gamma : [0,1] \to C,\qquad \gamma(t)=(1-t)x+ty. \end{align*} This formula will define a map into $C$ once we verify that $\gamma(t)\in C$ for every $t\in[0,1]$. [/step] [step:Use convexity to keep the segment inside $C$] Let $t\in[0,1]$. Since $x,y\in C$, since $t\in[0,1]$, and since $C$ is convex, the convex combination $(1-t)x+ty$ belongs to $C$. Hence $\gamma(t)\in C$ for every $t\in[0,1]$, so $\gamma : [0,1]\to C$ is well-defined. [guided] The only possible obstruction to using the straight line from $x$ to $y$ is that it might leave the subset $C$. Convexity is exactly the hypothesis that rules this out. By definition, a subset $C\subset \mathbb R^n$ is convex when, for every $a,b\in C$ and every scalar $\lambda\in[0,1]$, the point $(1-\lambda)a+\lambda b$ lies in $C$. We apply this definition with $a=x$, $b=y$, and $\lambda=t$. Since $x,y\in C$ and $t\in[0,1]$, convexity gives \begin{align*} (1-t)x+ty \in C. \end{align*} Therefore the formula $\gamma(t)=(1-t)x+ty$ really does define a map from $[0,1]$ into $C$, not merely a map from $[0,1]$ into the ambient space $\mathbb R^n$. [/guided] [/step] [step:Prove continuity in the ambient Euclidean space] Let $f : [0,1]\to\mathbb R^n$ be the same formula regarded as an ambient Euclidean map: \begin{align*} f(t)=(1-t)x+ty. \end{align*} For any $s,t\in[0,1]$, subtraction in $\mathbb R^n$ gives \begin{align*} f(t)-f(s)=(t-s)(y-x). \end{align*} Taking the Euclidean norm yields \begin{align*} |f(t)-f(s)|=|t-s|\,|y-x|. \end{align*} If $x=y$, then $f$ is constant and hence continuous. If $x\neq y$, let $\varepsilon>0$ and choose \begin{align*} \delta=\frac{\varepsilon}{|y-x|}. \end{align*} Whenever $s,t\in[0,1]$ and $|t-s|<\delta$, the previous estimate gives $|f(t)-f(s)|<\varepsilon$. Thus $f:[0,1]\to\mathbb R^n$ is continuous. [/step] [step:Pass continuity to the subspace $C$] We now prove that $\gamma:[0,1]\to C$ is continuous for the subspace topology on $C$. Let $O\subset C$ be open in the subspace topology. By definition of the subspace topology, there exists an [open set](/page/Open%20Set) $U\subset\mathbb R^n$ such that \begin{align*} O=U\cap C. \end{align*} Since $\gamma([0,1])\subset C$ and $f$ is the same point-valued map as $\gamma$ with codomain $\mathbb R^n$, we have \begin{align*} \gamma^{-1}(O)=f^{-1}(U). \end{align*} The set $f^{-1}(U)$ is open in $[0,1]$ because $f:[0,1]\to\mathbb R^n$ is continuous and $U$ is open in $\mathbb R^n$. Hence $\gamma^{-1}(O)$ is open in $[0,1]$. Since this holds for every open $O\subset C$, the map $\gamma:[0,1]\to C$ is continuous. [/step] [step:Check the endpoints and conclude path-connectedness] Evaluating the path at the endpoints gives \begin{align*} \gamma(0)=(1-0)x+0y=x \end{align*} and \begin{align*} \gamma(1)=(1-1)x+1y=y. \end{align*} Thus every pair of points $x,y\in C$ can be joined by a continuous path in $C$. Therefore $C$ is path-connected. [/step]