Von Neumann's minimax theorem

Weak duality. For any fixed $x_0 \in \Delta_m$ and $y_0 \in \Delta_n$: $\min_y x_0^\top A y \le x_0^\top A y_0 \le \max_x x^\top A y_0$. Taking $\max_{x_0}$ of the left side and $\min_{y_0}$ of the right side preserves the inequality: $\max_x \min_y x^\top A y \le \min_y \max_x x^\top A y$. This direction needs no randomization argument at all.

Reduction to a linear program. Since $y \mapsto x^\top A y$ is linear, its minimum over the simplex $\Delta_n$ is attained at a vertex, i.e. at some pure strategy $j$: $\min_y x^\top A y = \min_{j} \sum_{i=1}^m x_i A_{ij}$. So the row player's problem $\max_x \min_y x^\top A y$ is the linear program: maximize $v$ subject to $\sum_{i=1}^m A_{ij} x_i \ge v$ for every $j = 1, \dots, n$, and $x \in \Delta_m$.

The dual program. By the standard theory of linear programming duality, the dual of this LP is: minimize $w$ subject to $\sum_{j=1}^n A_{ij} y_j \le w$ for every $i = 1, \dots, m$, and $y \in \Delta_n$ — which is exactly the linear-programming formulation of the column player's problem $\min_y \max_x x^\top A y$.

Strong duality. Both the primal and dual feasible regions ($\Delta_m$ and $\Delta_n$) are nonempty and compact, so the linear program is feasible and bounded; the strong duality theorem for linear programming then guarantees the primal and dual optimal values coincide: $\max_x \min_y x^\top A y = \min_y \max_x x^\top A y$. Combined with weak duality, which already gave $\le$ in this direction, equality holds, proving the theorem.