I am self-studying some discrete geometry / convex analysis. Many descriptions of Caratheodory's Theorem for convex sets mention that Radon's Lemma can be used to simplify the proof, but I haven't seen it done. For reference, here is Radon's Lemma:
Lemma (Radon). Let $A \subset \mathbb{R}^d$ contain $d+2$ points. Then there exist two disjoint subsets $A_1, A_2 \subset A$ whose convex hulls have nonempty intersection.
I will attempt to prove:
Theorem (Caratheodory). Let $X \subset \mathbb{R}^d$. Then each point of $\mathrm{conv}(X)$ can be written as a convex combination of at most $d+1$ points in $X$.
Proof Attempt. Each $y \in \mathrm{conv}(X)$ is a convex combination $y = \sum_{k=1}^m \alpha_k x_k$ of finitely many points $x_1, \dots, x_m \in X$, where $\alpha_k > 0$ and $\sum_{k=1}^m \alpha_k = 1$. Assume $m \geq d+2$, otherwise we are done. Further assume towards contradiction that $m$ is minimal, that is, $y$ cannot be written as the convex combination of fewer than $m$ points from $X$.
Then, the points $x_1, \dots, x_m$ are affinely dependent, being $m \geq d+2$ points in $\mathbb{R}^d$; hence one point, say $x_m$, is an affine combination of the rest. Apply Radon's Lemma to the set $A = \{ y, x_1, \dots, x_{m-1} \}$, giving two sets $A_1, A_2 \subset A$ whose convex hulls have nonempty intersection....?
Is this the right idea? How might I continue?
You're in the right direction. By induction it's enough to prove that we can reduce a convex combination of $n+2$ points to a convex combination of $n+1$ points. Suppose we have a point $v\in\mathbb{R}^n$ and that it is a convex combination of $n+2$ points. This means that $$v=\sum_{i=1}^{n+2}\alpha_ix_i.$$ Where $x_i\in X, ~\sum_i\alpha_i=1$ and $\alpha_i\geq 0$. Since any $n+2$ points in $\mathbb{R}^n$ are affinely dependent, then we have $$\sum_{i=1}^{n+2}\beta_ix_i=0,$$ where $\sum_i\beta_i=1$.
Now, let $\gamma=\min\{\frac{\alpha_i}{\beta_i}:\beta_i>0\}>0$, without loss of generality, assume that $\gamma=\frac{a_{n+2}}{\beta_{n+2}}$ and pay attention to the following $$ v=v-\gamma\cdot 0=\sum_i^{n+2}(\alpha_i-\gamma\beta_i)x_i=\sum_i^{n+1}(\alpha_i-\gamma\beta_i)x_i, $$
clearly, every coefficient is positive and they sum to 1 which completes the proof.