Proof, [Lemma 4.49, Fol99]. Fix $\eps > 0$. Let $x \in X$, then for each $y \in Y$, $\bracs{g_{xy} > f - \eps}\in \cn_{X}(y)$. By compactness of $X$, there exists $Y_{x} \subset Y$ finite such that $X = \bigcup_{y \in Y_x}\bracs{g_{xy} > f - \eps}$. In which case, let $g_{x} = \bigvee_{y \in Y_x}g_{xy}$, then $g \in A$ and $g_{x}(x) = f(x)$ and $g > f - \eps$.

Similarly, for each $x \in X$, $\bracs{g_x < f + \eps}\in \cn_{X}(x)$. By compactness of $X$, there exists $X_{0} \subset X$ finite such that $X = \bigcup_{x \in X_0}\bracs{g_x < f + \eps}$. Let $g = \bigwedge_{x \in X_0}g_{x}$, then $g \in A$ and $\norm{f - g}_{u} < \eps$.$\square$