[proofplan] Let $B$ denote the set of all finite convex combinations of points of $A$. We first prove that $B$ is a convex subset of $V$ containing $A$, so the defining minimality of the convex hull gives $\operatorname{conv}(A) \subset B$. Conversely, every convex subset of $V$ containing $A$ contains all finite convex combinations of points of $A$, proved by induction on the number of points in the combination. Hence $B$ is contained in every convex superset of $A$, and therefore $B \subset \operatorname{conv}(A)$. [/proofplan]

[step:Define the finite convex-combination set and include $A$ in it] Define the subset $B \subset V$ by \begin{align*} B := \left\{\sum_{i=1}^m \lambda_i a_i : m \in \mathbb{N},\ a_i \in A,\ \lambda_i \geq 0\ \text{for all } i,\ \sum_{i=1}^m \lambda_i = 1\right\}. \end{align*} If $a \in A$, choose $m = 1$, $a_1 = a$, and $\lambda_1 = 1$. Then \begin{align*} a = \sum_{i=1}^1 \lambda_i a_i \in B. \end{align*} Thus $A \subset B$. If $A = \varnothing$, then no choice of $a_i \in A$ is possible, so $B = \varnothing$. [/step]

[step:Show that $B$ is convex by concatenating two convex combinations]We prove that $B$ is convex. If $B = \varnothing$, this is vacuous. Otherwise let $x,y \in B$ and let $t \in [0,1]$. By definition of $B$, there are $m,n \in \mathbb{N}$, points $a_1,\dots,a_m \in A$, points $c_1,\dots,c_n \in A$, coefficients $\lambda_1,\dots,\lambda_m \in [0,\infty)$, and coefficients $\mu_1,\dots,\mu_n \in [0,\infty)$ such that \begin{align*} x = \sum_{i=1}^m \lambda_i a_i, \end{align*} \begin{align*} y = \sum_{j=1}^n \mu_j c_j, \end{align*} \begin{align*} \sum_{i=1}^m \lambda_i = 1, \end{align*} and \begin{align*} \sum_{j=1}^n \mu_j = 1. \end{align*} Define points $b_1,\dots,b_{m+n} \in A$ by $b_i = a_i$ for $1 \leq i \leq m$ and $b_{m+j} = c_j$ for $1 \leq j \leq n$. Define coefficients $\theta_1,\dots,\theta_{m+n} \in [0,\infty)$ by $\theta_i = t\lambda_i$ for $1 \leq i \leq m$ and $\theta_{m+j} = (1-t)\mu_j$ for $1 \leq j \leq n$. Since $t \in [0,1]$, all $\theta_k$ are nonnegative, and \begin{align*} \sum_{k=1}^{m+n} \theta_k = t\sum_{i=1}^m \lambda_i + (1-t)\sum_{j=1}^n \mu_j = t + (1-t) = 1. \end{align*} Moreover, \begin{align*} tx + (1-t)y = \sum_{k=1}^{m+n} \theta_k b_k. \end{align*} Therefore $tx + (1-t)y \in B$. Hence $B$ is convex.[/step]

[guided]We need to check the defining closure property for convexity: whenever $x,y \in B$ and $t \in [0,1]$, the point $tx + (1-t)y$ must again belong to $B$. Since $x$ and $y$ are elements of $B$, each is already written as a finite convex combination of points of $A$. Thus there are $m,n \in \mathbb{N}$, points $a_1,\dots,a_m \in A$ and $c_1,\dots,c_n \in A$, and coefficients $\lambda_1,\dots,\lambda_m,\mu_1,\dots,\mu_n \in [0,\infty)$ such that \begin{align*} x = \sum_{i=1}^m \lambda_i a_i, \end{align*} \begin{align*} y = \sum_{j=1}^n \mu_j c_j, \end{align*} with \begin{align*} \sum_{i=1}^m \lambda_i = 1 \end{align*} and \begin{align*} \sum_{j=1}^n \mu_j = 1. \end{align*} The natural way to express $tx + (1-t)y$ as one convex combination is to concatenate the two lists of points and multiply the first list of coefficients by $t$ and the second list by $1-t$. Define $b_1,\dots,b_{m+n} \in A$ by $b_i = a_i$ for $1 \leq i \leq m$ and $b_{m+j} = c_j$ for $1 \leq j \leq n$. Define coefficients $\theta_1,\dots,\theta_{m+n}$ by $\theta_i = t\lambda_i$ for $1 \leq i \leq m$ and $\theta_{m+j} = (1-t)\mu_j$ for $1 \leq j \leq n$. Because $t \in [0,1]$, both $t$ and $1-t$ are nonnegative, so every coefficient $\theta_k$ is nonnegative. The sum of the coefficients is \begin{align*} \sum_{k=1}^{m+n} \theta_k = t\sum_{i=1}^m \lambda_i + (1-t)\sum_{j=1}^n \mu_j = t + (1-t) = 1. \end{align*} Finally, the vector-space operations give \begin{align*} tx + (1-t)y = t\sum_{i=1}^m \lambda_i a_i + (1-t)\sum_{j=1}^n \mu_j c_j = \sum_{k=1}^{m+n} \theta_k b_k. \end{align*} This is exactly the required form of an element of $B$. Therefore $B$ is closed under convex combinations of two points, so $B$ is convex. If $B = \varnothing$, the same convexity condition is vacuously satisfied because there are no points $x,y \in B$ to test.[/guided]

[step:Use minimality of the convex hull to get $\operatorname{conv}(A) \subset B$] The set $B$ is convex and contains $A$. By the definition of $\operatorname{conv}(A)$ as the smallest convex subset of $V$ containing $A$, equivalently the intersection of all convex subsets of $V$ containing $A$, we obtain \begin{align*} \operatorname{conv}(A) \subset B. \end{align*} When $A = \varnothing$, this also agrees with the standard convention that $\varnothing$ is convex, so $\operatorname{conv}(\varnothing)=\varnothing=B$. [/step]

[step:Show that every convex superset of $A$ contains $B$] Let $C \subset V$ be any convex subset such that $A \subset C$. We prove by induction on $m \in \mathbb{N}$ that every convex combination of $m$ points of $A$ belongs to $C$. For $m=1$, if $a_1 \in A$ and $\lambda_1 \geq 0$ with $\lambda_1=1$, then \begin{align*} \sum_{i=1}^1 \lambda_i a_i = a_1 \in A \subset C. \end{align*} Assume the assertion holds for all convex combinations of $m-1$ points of $A$, where $m \geq 2$. Let $a_1,\dots,a_m \in A$ and let $\lambda_1,\dots,\lambda_m \in [0,\infty)$ satisfy \begin{align*} \sum_{i=1}^m \lambda_i = 1. \end{align*} Set \begin{align*} s := \sum_{i=1}^{m-1} \lambda_i = 1-\lambda_m. \end{align*} If $s=0$, then $\lambda_m=1$ and $\lambda_i=0$ for $1 \leq i \leq m-1$, so \begin{align*} \sum_{i=1}^m \lambda_i a_i = a_m \in A \subset C. \end{align*} If $s>0$, define coefficients $\alpha_1,\dots,\alpha_{m-1} \in [0,\infty)$ by \begin{align*} \alpha_i := \frac{\lambda_i}{s}. \end{align*} Then \begin{align*} \sum_{i=1}^{m-1} \alpha_i = 1. \end{align*} By the induction hypothesis, \begin{align*} z := \sum_{i=1}^{m-1} \alpha_i a_i \end{align*} belongs to $C$. Also $a_m \in A \subset C$. Since $C$ is convex and $s \in [0,1]$, the point \begin{align*} sz + (1-s)a_m \end{align*} belongs to $C$. Substituting the definition of $z$ and using $1-s=\lambda_m$, we get \begin{align*} sz + (1-s)a_m = \sum_{i=1}^{m-1} \lambda_i a_i + \lambda_m a_m = \sum_{i=1}^m \lambda_i a_i. \end{align*} Thus every convex combination of $m$ points of $A$ lies in $C$. The induction is complete, and therefore $B \subset C$. [/step]

[step:Intersect over all convex supersets of $A$ to obtain the reverse inclusion] Since $C$ was an arbitrary convex subset of $V$ containing $A$, the preceding step shows that $B$ is contained in every convex subset of $V$ containing $A$. Taking the intersection over all such convex subsets gives \begin{align*} B \subset \operatorname{conv}(A). \end{align*} Combining this with $\operatorname{conv}(A) \subset B$ yields \begin{align*} \operatorname{conv}(A) = B. \end{align*} This is precisely the asserted characterization of $\operatorname{conv}(A)$ by finite convex combinations of points of $A$. [/step]