The Erdős–Ko–Rado theorem

Idea (Katona). Count pairs (circular arrangement, arc of the arrangement belonging to $\mathcal F$).

Step 1: at most $k$ arcs. Fix an arrangement of $1,\dots,n$ at positions $0,\dots,n-1$ of a circle, and call $I_t$ the arc of $k$ consecutive positions starting at $t$. Suppose $I_0\in\mathcal F$. Any other arc meeting $I_0$ starts at a position $t\in\{1,\dots,k-1\}$ or $t\in\{-1,\dots,-(k-1)\}$. For $i=1,\dots,k-1$ the two arcs $I_i$ and $I_{i-k}$ are consecutive blocks of $k$ positions, hence disjoint because $2k\le n$. Being intersecting, $\mathcal F$ contains at most one of $I_i$, $I_{i-k}$. So at most $1+(k-1)=k$ arcs of the arrangement are in $\mathcal F$.

Step 2: double counting. Over all $n!$ arrangements, the number $N$ of pairs (arrangement, arc in $\mathcal F$) satisfies $N\le k\cdot n!$. On the other hand, a given set $A\in\mathcal F$ is an arc in exactly $n\,k!\,(n-k)!$ arrangements. So
$$N=|\mathcal F|\cdot n\,k!\,(n-k)!\le k\cdot n!\ \Longrightarrow\ |\mathcal F|\le\frac kn\binom nk=\binom{n-1}{k-1}.$$
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Sharpness. All $k$-sets containing the element $1$ form an intersecting family of size $\binom{n-1}{k-1}$.

Lesson. A cyclic arrangement turns a global set-system problem into a local one; then double count.