Show that any continuous function $f:[0,1]^2 \to \mathbb R$ and anny $\epsilon > 0$ there exists a natural number $n$ and functions $g_1,...,g_n, h_1,...,h_n \in C([0,1]) $ such that $$\left|f(x,y) - \sum_{k=1}^n g_k(x)h_k(y)\right| < \epsilon$$ for all $(x,y) \in [0,1]^2$.
I am really unsure how to approach this problem. It seems that I might somehow have to fix $x$ and then work something out and then fix $y$ and then work something out, using triangle inequality. I was thinking I might be able to use the Berstein polynomials, but I am not exactly sure how that might work out. My ad-hoc idea is that, in a small neighborhood around some particular $x_0$, we can view $x$ as "essentially constant" since it doesn't vary much. Hence we can view the function $f(x,y)$ as almost as a single valued function $f(x_0,y)$, which can be approximated by Berstein polynomials with $y$ as its variable. But this idea is dangerous in a sense that (1) the "constants" we take (in terms of $x_0$ ) cannot be guaranteed to make the supposed $g_k(x)$ continuous, and (2) this idea is in general very sketchy without enough details, especially as the Berstein polynomial approximation is already somewhat complex in single variable. Is there any other way to approach this problem? My idea seems to be pretty complicated.
I do not know of an elementary proof but this follows easily from Stone-Weierstrass Theorem. Finite sums of the type $\sum_{i=1} ^{n} g_i (x) h_i (y)$ form an algebra and contains constants; it separates points: $(a,b) \neq (c,d)$ implies either $a \neq c$ or $c \neq d$. In the first case $x$ separates the two points and in the second case $y$ does.