[proofplan] To prove convexity of the intersection, we take two arbitrary points in the intersection and an arbitrary convex parameter. Membership in the intersection means membership in every set $C_i$. Since each $C_i$ is convex, the convex combination lies in every $C_i$, and hence lies in their intersection. [/proofplan] [step:Choose two points in the intersection and a convex parameter] Let $x, y \in C$, and let $\lambda \in [0,1]$. We must prove that $(1-\lambda)x + \lambda y \in C$. [/step] [step:Use convexity of each member of the family] Fix an arbitrary index $i \in I$. Since $x, y \in C$, the definition of $C$ gives $x \in C_i$ and $y \in C_i$. Since $C_i$ is convex and $\lambda \in [0,1]$, we have \begin{align*} (1-\lambda)x + \lambda y \in C_i. \end{align*} [/step] [step:Conclude membership in the full intersection] The index $i \in I$ was arbitrary, so \begin{align*} (1-\lambda)x + \lambda y \in C_i \end{align*} for every $i \in I$. By the definition of intersection, \begin{align*} (1-\lambda)x + \lambda y \in \bigcap_{i \in I} C_i = C. \end{align*} Therefore $C$ is convex. [/step]