[proofplan]
Set $U := \bigcup_{i \in I} K_i$ and $C := \operatorname{conv}(U)$. The inclusion $C^\circ \subset \bigcap_{i \in I} K_i^\circ$ follows because each $K_i$ is contained in $C$. For the reverse inclusion, a point of $C$ is a finite convex combination of points from $U$, and the inequality defining the polar is preserved under finite convex averaging by linearity of the Euclidean inner product in the first variable.
[/proofplan]
[step:Restrict the polar inequality from the convex hull to each generating set]
Define
\begin{align*}
U := \bigcup_{i \in I} K_i,
\qquad
C := \operatorname{conv}(U).
\end{align*}
Let $y \in C^\circ$. Fix $i \in I$ and $x \in K_i$. Since $K_i \subset U \subset C$, the definition of $C^\circ$ gives
\begin{align*}
x \cdot y \leq 1.
\end{align*}
Therefore $y \in K_i^\circ$. Since $i \in I$ was arbitrary, $y \in \bigcap_{i \in I} K_i^\circ$. Hence
\begin{align*}
C^\circ \subset \bigcap_{i \in I} K_i^\circ.
\end{align*}
[/step]
[step:Average the inequalities over an arbitrary finite convex combination]
Let $y \in \bigcap_{i \in I} K_i^\circ$. We prove that $y \in C^\circ$.
Fix $x \in C$. By the definition of convex hull, there exist $m \in \mathbb{N}$, points $x_1,\dots,x_m \in U$, and scalars $\lambda_1,\dots,\lambda_m \in [0,\infty)$ such that
\begin{align*}
\sum_{j=1}^{m} \lambda_j = 1,
\qquad
x = \sum_{j=1}^{m} \lambda_j x_j.
\end{align*}
For each $j \in \{1,\dots,m\}$, since $x_j \in U$, there exists $i_j \in I$ such that $x_j \in K_{i_j}$. Because $y \in \bigcap_{i \in I} K_i^\circ$, we have $y \in K_{i_j}^\circ$, and hence
\begin{align*}
x_j \cdot y \leq 1
\end{align*}
for every $j \in \{1,\dots,m\}$.
Using linearity of the map
\begin{align*}
\ell_y: \mathbb{R}^n &\to \mathbb{R} \\
z &\mapsto z \cdot y
\end{align*}
and the non-negativity of the coefficients $\lambda_j$, we obtain
\begin{align*}
x \cdot y
&=
\left(\sum_{j=1}^{m} \lambda_j x_j\right) \cdot y \\
&=
\sum_{j=1}^{m} \lambda_j (x_j \cdot y) \\
&\leq
\sum_{j=1}^{m} \lambda_j \\
&=
1.
\end{align*}
Thus $x \cdot y \leq 1$ for every $x \in C$, so $y \in C^\circ$. Therefore
\begin{align*}
\bigcap_{i \in I} K_i^\circ \subset C^\circ.
\end{align*}
[guided]
Let $y \in \bigcap_{i \in I} K_i^\circ$. The meaning of this assumption is that the same vector $y$ satisfies the polar inequality on every set $K_i$: for each $i \in I$ and every $z \in K_i$,
\begin{align*}
z \cdot y \leq 1.
\end{align*}
We must show that the inequality also holds on the whole convex hull $C$.
Take an arbitrary point $x \in C$. By the definition of $C = \operatorname{conv}(U)$, the point $x$ is a finite convex combination of points from $U$. Thus there exist $m \in \mathbb{N}$, points $x_1,\dots,x_m \in U$, and coefficients $\lambda_1,\dots,\lambda_m \in [0,\infty)$ such that
\begin{align*}
\sum_{j=1}^{m} \lambda_j = 1,
\qquad
x = \sum_{j=1}^{m} \lambda_j x_j.
\end{align*}
For each $j$, the condition $x_j \in U = \bigcup_{i \in I}K_i$ means that there is an index $i_j \in I$ with $x_j \in K_{i_j}$. Since $y$ lies in every polar $K_i^\circ$, in particular $y \in K_{i_j}^\circ$, so
\begin{align*}
x_j \cdot y \leq 1
\end{align*}
for every $j \in \{1,\dots,m\}$.
Now define the linear functional determined by $y$:
\begin{align*}
\ell_y: \mathbb{R}^n &\to \mathbb{R} \\
z &\mapsto z \cdot y.
\end{align*}
Linearity is the mechanism that transfers the pointwise inequalities from the generators $x_j$ to their convex average. Indeed,
\begin{align*}
x \cdot y
&=
\ell_y(x) \\
&=
\ell_y\left(\sum_{j=1}^{m} \lambda_j x_j\right) \\
&=
\sum_{j=1}^{m} \lambda_j \ell_y(x_j) \\
&=
\sum_{j=1}^{m} \lambda_j (x_j \cdot y) \\
&\leq
\sum_{j=1}^{m} \lambda_j \\
&=
1.
\end{align*}
The inequality step uses both facts that define a convex combination: each $\lambda_j$ is non-negative, and the coefficients sum to $1$. Therefore $x \cdot y \leq 1$ for every $x \in C$, which is precisely $y \in C^\circ$.
[/guided]
[/step]
[step:Combine the two inclusions]
The first inclusion gives
\begin{align*}
C^\circ \subset \bigcap_{i \in I} K_i^\circ,
\end{align*}
and the second gives
\begin{align*}
\bigcap_{i \in I} K_i^\circ \subset C^\circ.
\end{align*}
Therefore
\begin{align*}
C^\circ = \bigcap_{i \in I} K_i^\circ.
\end{align*}
Substituting $C = \operatorname{conv}\left(\bigcup_{i \in I}K_i\right)$ yields
\begin{align*}
\left(\operatorname{conv}\left(\bigcup_{i \in I}K_i\right)\right)^\circ
=
\bigcap_{i \in I} K_i^\circ,
\end{align*}
as required.
[/step]
