en.wikipedia.org/wiki/Helly
We prove the finite version, using Radon's theorem as in the proof by Radon(1921). The infinite version then follows by the finite intersection property characterization of compact space: a collection of closed subsets of a compact space has a non-empty intersection if and only if every finite subcollection has a non-empty intersection (once you fix a single set, the intersection of all others with it are closed subsets of a fixed compact space).
The proof is by induction:
Base case: Let $n=d+2$. By our assumptions, for every $j=1,⋯,n$ there is a point $x_j$ that is in the common intersection of all $X_i$ with the possible exception of $X_j$. Now we apply Radon's theorem to the set $A=x_1,...,x_n$, which furnishes us with disjoint subsets $A_1,A_2$ of $A$ such that the convex hull of $A_1$ intersects the convex hull of $A_2$. Suppose that $p$ is a point in the intersection of these two convex hulls. We claim that
$$p\in\bigcap_{j=1}^n X_j.$$
Indeed, consider any $j∈\{1,...,n\}$. We shall prove that $p∈X_j$. Note that the only element of $A$ that may not be in $X_j$ is $x_j$. If $x_j∈A_1$, then $x_j∉A_2$, and therefore $X_j⊃A_2$. Since $X_j$ is convex, it then also contains the convex hull of $A_2$ and therefore also $p∈X_j$. Likewise, if $x_j∉A_1$, then $X_j⊃A_1$, and by the same reasoning $p∈X_j$. Since $p$ is in every $X_j$, it must also be in the intersection.
Above, we have assumed that the points $x_1,...,x_n$ are all distinct. If this is not the case, say $x_i=x_k$ for some $i≠k$, then $x_i$ is in every one of the sets $X_j$, and again we conclude that the intersection is nonempty. This completes the proof in the case $n=d+2$.
Inductive Step: Suppose $n>d+2$ and that the statement is true for $n−1$. The argument above shows that any subcollection of $d+2$ sets will have nonempty intersection. We may then consider the collection where we replace the two sets $X_{n−1}$ and $X_n$ with the single set $X_{n−1}∩ X_n$. In this new collection, every subcollection of $d+1$ sets will have nonempty intersection. The inductive hypothesis therefore applies, and shows that this new collection has nonempty intersection. This implies the same for the original collection, and completes the proof.
The Art of Mathematics: Coffee Time in Memphis, page 23-24
28. Convexity and Intersecting Simplices
(i) Let $X=\left\{\mathbf{x}_{1}, \mathbf{x}_{2}, \ldots, \mathbf{x}_{n+2}\right\}$ be a set of $n+2$ points in $\mathbb{R}^{n}$ and, for a non-empty subset $I$ of $\{1, \ldots, n+2\}$, let $X(I)$ be the convex hull of the points $\mathbf{x}_{i}, i \in I$. Show that there are disjoint sets $I, J$ such that $X(I) \cap$ $X(J) \neq \varnothing .$
(ii) The convex hull $\operatorname{conv} X$ of a set $X \subset \mathbb{R}^{n}$ is the smallest convex set containing $X$ : the intersection of all convex sets containing it. Equivalently,
$$
\operatorname{conv} X=\left\{\sum_{i=1}^{k} \lambda_{i} \mathbf{x}_{i}: \mathbf{x}_{i} \in X, \lambda_{i} \geq 0, \sum_{i=1}^{k} \lambda_{i}=1, k=1,2, \ldots\right\}
$$
Show that in the sums above we need not take more than $n+1$ terms, i.e., the convex hull of a set $X \subset \mathbb{R}^{n}$ is
$$
\operatorname{conv} X=\left\{\sum_{i=1}^{n+1} \lambda_{i} \mathbf{x}_{i}: \mathbf{x}_{i} \in X, \lambda_{i} \geq 0, \sum_{i=1}^{n+1} \lambda_{i}=1\right\}
$$ 29. Intersecting Convex Sets. Let $\mathcal{C}$ be a finite family of convex sets in $\mathbb{R}^{n}$ such that for $k \leq n+1$ any $k$ members of $\mathcal{C}$ have a non-empty intersection. Show that the intersection of all members of $\mathcal{C}$ is non-empty. 30. Judicious Partitions of Points. Let $P_{1}, \ldots, P_{n}$ be $n$ points in the plane. Show that there is a point $P$ such that every line through $P$ has at least $n / 3$ points $P_{i}$ in each of the two closed half-planes it determines.
