Let $V$ be a compact Hausdorff space and $F\colon V \times V \to \mathbb{R}$ be a continuous function. I want to show that, for every $\epsilon > 0$ there is a partition $P$ of $V$ into finitely many sets such that the variation of $F$ inside each set $S \times T$ with $S$, $T \in P$ is bounded by $\epsilon$.
If $V$ is a metric space, then I can prove this, otherwise not. Am I missing something? There is a rather complicated theory of uniform spaces (https://en.wikipedia.org/wiki/Uniform_space), but I don't know if I really need to use it. Any hints?
I assume the following definition of "variation of $F$ in $S\times T$": $$\text{Var}_{S\times T}(F) := \left[\sup_{(x,y) \in S\times T} F(x,y) \right] -\left[\inf_{(x,y) \in S\times T} F(x,y) \right].$$
Fix $(x,y) \in V\times V$. Since $F$ is continuous, for any $\epsilon > 0$, we can find an open rectangle $S_{x,y} \times T_{x,y} \subset V\times V$ such that for any $(z,w) \in S_{x,y} \times T_{x,y}$, we have $$\lvert F(x,y) - F(z,w) \rvert < \epsilon/3.$$ But then for any $(z_1,w_1), (z_2,w_2) \in S_{x,y} \times T_{x,y}$, by the triangle inequality we have $$\lvert F(z_1,w_1) - F(z_2,w_2) \rvert < 2\epsilon/3$$ and so $$\text{Var}_{S_{x,y} \times T_{x,y}}(F) \le 2\epsilon / 3 < \epsilon.$$ Now the collection of sets $$\{S_{x,y} \times T_{x,y} \}_{(x,y) \in V\times V}$$ form an open cover of $V\times V$ so by compactness we can find a finite subcover (which we re-index) $$\{ S_{i} \times T_i\}_{i=1}^N.$$ Finally, replace $S_i \times T_i$ with $S'_i \times T'_i = S_i \times T_i \setminus \cup^N_{j=i+1} S_j \times T_j$ and the sets $$\{S_i' \times T_i' \}_{i=1}^N$$ form the desired partition. They are disjoint by definition, and the variation of $F$ is small on each rectangle $S'_i \times T_i'$ since each of these rectangles is a subset of some rectangle $S_{x,y} \times T_{x,y}$ which were constructed to admit small variation.