 Solution

Let $P$ be the row payoff matrix and $Q=P^T$. For $x$ in the probability simplex $\Delta$, write $u(x)=x^{T}Px$ and define the symmetric Nash gain mapThe map is a continuous function, has nonnegative coordinates, and sums to one. The Brouwer fixed-point theorem states that every continuous self-map of a nonempty finite-dimensional compact convex set has a fixed point. Apply it to $T:\Delta \to \Delta$, and let $x=T(x)$.

Set $G={\sum }_i{g}_i(x)$. The fixed-point equation gives $g_i(x)=G{x}_i$. If $G>0$, every positive $x_i$ has strictly positive gain, so $(Px{)}_i-u(x)=g_i(x)=G{x}_i$. Buta contradiction. Thus $G=0$ and every pure payoff $(Px{)}_i$ is at most $u(x)$. Since their $x$-weighted average equals $u(x)$, every supported action attains that maximum. Consequently $x$ is a best response to itself.

The column player's payoff vector against $x$ is $Q^{T}x=Px$, so exactly the same inequalities establish its best response. ThereforeThis supplies the whole fixed-point construction and fixed-point-to-equilibrium argument, rather than assuming Nash's theorem as a black box.