Prove that there is $a \in X$ s.t $a \in \underset{z\in X}{\operatorname{arg max}} f(z, a)$

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I'm proving that a certain strategic game has a symmetric Nash equilibrium. The problem boils down to below theorem:

Let $X$ be nonempty convex compact and $f: X^2 \to \mathbb R$ continuous. Moreover, $f$ is quasi-concave w.r.t the first coordinate. Then there is $a \in X$ such that $$a \in \underset{z\in X}{\operatorname{arg max}} f(z, a)$$

I've tried to use below Kakutani-Fan fixed point theorem from my lecture note:

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Could you please verify if my proof looks fine or contains logical gaps/errors? Thank you so much for your help!


My attempt:

We define a set-valued function $$g: X \to \mathcal P(X), \quad x \mapsto \underset{z\in X}{\operatorname{arg max}} f(z,x)$$

Because $f$ is quasi-concave w.r.t the first coordinate, $g(x) = \underset{z\in X}{\operatorname{arg max}} f(z,x)$ is convex. Because $f$ is continuous, $g(x) = \underset{z\in X}{\operatorname{arg max}} f(z,x)$ is nonempty closed. Because $g(x) \subseteq X$ which is compact, $g(x)$ is compact.

Let $A = \{(x,y) \in X^2 \mid y \in g(x)\}$ and $(x_n,y_n)_{n \in \mathbb N}$ be a sequence in $A$ such that $(x_n,y_n) \to (x^*,y^*) \in X^2$, and thus $\forall z \in X: \lim_{n \to \infty} f(y_n, x_n) \ge \lim_{n \to \infty} f(z,x_n)$ for all $n \in \mathbb N$. By the continuity of $f$, we get $\forall z \in X: f(\lim_{n \to \infty} y_n, \lim_{n \to \infty} x_n) \ge f(z, \lim_{n \to \infty} x_n)$ for all $n \in \mathbb N$. As a result, $\forall z \in X: f(y^*, x^*) \ge f(z, x^*)$ for all $n \in \mathbb N$. So $(x^*,y^*) \in A$ and hence $A$ is closed.

By Kakutani-Fan fixed point theorem, $g$ has a fixed point, i.e., there is $a \in X$ such that $$a \in \underset{z\in X}{\operatorname{arg max}} f(z, a)$$