= Solution
Let $P$ be the row payoff matrix and $Q=P^T$. For $x$ in the <probability simplex> $\Delta$, write $u(x)=x^TPx$ and define the <symmetric Nash gain map>
$$
g_i(x)=\max\{0,(Px)_i-u(x)\},\qquad
T_i(x)=\frac{x_i+g_i(x)}{1+\sum_jg_j(x)}.
$$
The 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 $g_i(x)=Gx_i$. If $G>0$, every positive $x_i$ has strictly positive gain, so $(Px)_i-u(x)=g_i(x)=Gx_i$. But
$$
0=\sum_i x_i((Px)_i-u(x))=G\sum_i x_i^2>0,
$$
a 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 $Q^Tx=Px$, so exactly the same inequalities establish its <best response>. Therefore
$$
\boxed{(x,x)\text{ is a symmetric Nash equilibrium}.}
$$
This supplies the whole fixed-point construction and fixed-point-to-equilibrium argument, rather than assuming <Nash's theorem> as a black box.
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