= Solution
The set $E=\Delta_2\times\Delta_2$ is a compact convex square. Define
$$
u_i(p,q)=\max\{0,A(e_i,q)-A(p,q)\},
\qquad
p'=\frac{p+u(p,q)}{1+u_1(p,q)+u_2(p,q)}.
$$
Similarly, set
$$
v_j(p,q)=\max\{0,B(p,e_j)-B(p,q)\},
\qquad
q'=\frac{q+v(p,q)}{1+v_1(p,q)+v_2(p,q)}.
$$
These formulae define a continuous self-map $(p,q)\mapsto(p',q')$ of $E$. The square is homeomorphic to a closed disc, so the <Brouwer fixed-point theorem> gives a fixed point $(p^*,q^*)$.
Let $U=u_1+u_2$ at that fixed point. From $p'=p$,
$$
u_i=p_iU.
$$
If $U>0$, every $i$ with $p_i>0$ has $A(e_i,q)-A(p,q)>0$. This contradicts bilinearity, because
$$
\sum_i p_i\bigl(A(e_i,q)-A(p,q)\bigr)=0.
$$
Hence $U=0$, all $u_i=0$, and
$$
A(p^*,q^*)\geq A(e_i,q^*)
$$
for both pure strategies. By linearity this holds for every mixed strategy $p$. The same argument with $v_j$ gives
$$
B(p^*,q^*)\geq B(p^*,q)
$$
for every $q$. Therefore
$$
\boxed{
\forall(p,q)\in E,\quad
A(p^*,q^*)\geq A(p,q^*),
\qquad
B(p^*,q^*)\geq B(p^*,q)
}.
$$
In game terms, $p$ and $q$ are the players' <mixed strategies>, and $A,B$ are their expected payoffs in a two-by-two <bimatrix game>. The inequalities say that neither player can improve unilaterally, so $(p^*,q^*)$ is a <Nash equilibrium>. This is the <Brouwer proof of Nash equilibrium for a two-by-two game>.
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