Nash's existence theorem

Let $\Delta = \Delta_1 \times \cdots \times \Delta_n$, the product of the players' mixed-strategy simplices — a compact, convex subset of Euclidean space. For $x \in \Delta$, player $i$, and pure strategy $j \in S_i$, define the gain function $g_{ij}(x) = \max\big(0,\, u_i(s_{ij}, x_{-i}) - u_i(x)\big)$, the payoff player $i$ would gain by switching entirely to pure strategy $j$, or $0$ if that switch would not help. Define $f : \Delta \to \Delta$ by $f_i(x)_j = \dfrac{x_{ij} + g_{ij}(x)}{1 + \sum_{k \in S_i} g_{ik}(x)}$; since $u_i$ is continuous (indeed multilinear) in $x$, $g_{ij}$ is continuous, so $f$ is continuous, and each $f_i(x)$ is again a probability vector by construction.

By Brouwer's fixed-point theorem, the continuous map $f$ on the compact convex set $\Delta$ has a fixed point $x^\star$ with $f(x^\star) = x^\star$. Write $S_i := \sum_{k \in S_i} g_{ik}(x^\star)$; the fixed-point equation for each $(i,j)$ reads $x_{ij}^\star (1 + S_i) = x_{ij}^\star + g_{ij}(x^\star)$, i.e. $x_{ij}^\star S_i = g_{ij}(x^\star)$.

Fix a player $i$ and let $d_{ij} := u_i(s_{ij}, x_{-i}^\star) - u_i(x^\star)$, so $g_{ij}(x^\star) = \max(0, d_{ij})$. Multiply the fixed-point equation by $d_{ij}$ and sum over $j \in S_i$: on the left, $S_i \sum_j x_{ij}^\star d_{ij} = S_i\left(\sum_j x_{ij}^\star u_i(s_{ij}, x_{-i}^\star) - u_i(x^\star)\sum_j x_{ij}^\star\right) = S_i(u_i(x^\star) - u_i(x^\star)) = 0$, using $\sum_j x_{ij}^\star u_i(s_{ij}, x_{-i}^\star) = u_i(x^\star)$ and $\sum_j x_{ij}^\star = 1$.

On the right, $\sum_j g_{ij}(x^\star)\, d_{ij} = \sum_j \max(0,d_{ij})\, d_{ij} = \sum_{j:\, d_{ij} > 0} d_{ij}^2 \ge 0$. Equating both sides, $\sum_{j:\, d_{ij}>0} d_{ij}^2 = 0$, a sum of squares equal to zero, so every term vanishes: there is no $j$ with $d_{ij} > 0$. Since $i$ was arbitrary, $u_i(s_{ij}, x_{-i}^\star) \le u_i(x^\star)$ for every player $i$ and every pure strategy $j$ — precisely the Nash equilibrium condition. Hence $x^\star$ is a Nash equilibrium.