[proofplan] We prove the forward implication by induction on the number of points in the convex combination. The induction step separates the last coefficient: if it is $1$, the combination is just the last point; otherwise the first $m$ coefficients are renormalized to form a convex combination of the first $m$ points. The reverse implication follows by applying the finite-combination property with $m=2$, which is exactly the defining condition for convexity. [/proofplan]

[step:Prove finite convex combinations belong to a convex set by induction] Assume $C$ is convex. For each $m \in \mathbb{N}$, let $P_m$ be the assertion that for all points $x_1,\dots,x_m \in C$ and all coefficients $\lambda_1,\dots,\lambda_m \in [0,1]$ with \begin{align*} \sum_{i=1}^m \lambda_i = 1, \end{align*} one has \begin{align*} \sum_{i=1}^m \lambda_i x_i \in C. \end{align*} For $m=1$, the condition $\lambda_1=1$ gives \begin{align*} \sum_{i=1}^1 \lambda_i x_i = x_1 \in C, \end{align*} so $P_1$ holds. Assume $P_m$ holds for some $m \in \mathbb{N}$. Let $x_1,\dots,x_m,x_{m+1} \in C$ and let $\lambda_1,\dots,\lambda_m,\lambda_{m+1} \in [0,1]$ satisfy \begin{align*} \sum_{i=1}^{m+1} \lambda_i = 1. \end{align*} If $\lambda_{m+1}=1$, then $\sum_{i=1}^m \lambda_i=0$. Since each $\lambda_i \geq 0$, this implies $\lambda_i=0$ for every $1 \leq i \leq m$, and therefore \begin{align*} \sum_{i=1}^{m+1} \lambda_i x_i = x_{m+1} \in C. \end{align*} Now suppose $\lambda_{m+1}<1$. Define coefficients $\mu_1,\dots,\mu_m \in [0,1]$ by \begin{align*} \mu_i := \frac{\lambda_i}{1-\lambda_{m+1}}, \qquad 1 \leq i \leq m. \end{align*} The denominator is positive, and \begin{align*} \sum_{i=1}^m \mu_i = \frac{1}{1-\lambda_{m+1}}\sum_{i=1}^m \lambda_i = \frac{1-\lambda_{m+1}}{1-\lambda_{m+1}} = 1. \end{align*} By the induction hypothesis $P_m$, the point \begin{align*} y := \sum_{i=1}^m \mu_i x_i \end{align*} belongs to $C$. Since $y \in C$, $x_{m+1} \in C$, and $C$ is convex, the convex combination \begin{align*} (1-\lambda_{m+1})y + \lambda_{m+1}x_{m+1} \end{align*} belongs to $C$. Substituting the definition of $y$ gives \begin{align*} (1-\lambda_{m+1})y + \lambda_{m+1}x_{m+1} &= (1-\lambda_{m+1})\sum_{i=1}^m \frac{\lambda_i}{1-\lambda_{m+1}}x_i + \lambda_{m+1}x_{m+1} \\ &= \sum_{i=1}^{m+1} \lambda_i x_i. \end{align*} Thus $P_{m+1}$ holds. By induction, $P_m$ holds for every $m \in \mathbb{N}$. [/step]

[step:Recover convexity from the two-point case] Assume that every finite convex combination of points of $C$ belongs to $C$. Let $x,y \in C$ and let $t \in [0,1]$. Apply the assumed property with $m=2$, with points $x_1:=x$ and $x_2:=y$, and with coefficients \begin{align*} \lambda_1 := 1-t, \qquad \lambda_2 := t. \end{align*} Then $\lambda_1,\lambda_2 \in [0,1]$ and \begin{align*} \lambda_1+\lambda_2 = (1-t)+t = 1. \end{align*} Therefore \begin{align*} (1-t)x + ty = \lambda_1 x_1 + \lambda_2 x_2 \in C. \end{align*} Since this holds for all $x,y \in C$ and all $t \in [0,1]$, the set $C$ is convex. This proves the equivalence. [/step]