# Prove $A$ is dense in $C([0,1]\times[0,1])$

Given $$A=\left\{\sum^n_{k=0}f_k(x)g_k(y) : \ n \in \mathbb{Z}^+, \ f_k, g_k\in C[0,1]\right\}.$$

I am trying to use the Stone-Weierstrass Theorem to prove that $$A$$ is dense in $$C([0,1]\times[0,1])$$.

It is easy to see that $$A$$ is an algebra. I know $$A$$ vanishes nowhere, but I am confused about how to prove $$A$$ separates points. I know according to the Weierstrass Approximation Theorem, for all $$\epsilon>0$$ there exists a polynomial $$p$$ such that $$|p(x,y)-f(x,y)|<\epsilon$$, when $$f\in C([0,1]\times[0,1])$$, and I know polynomials separate points, and I have proved that all polynomials are in $$A$$.

My question is, how to prove $$A$$ separates points?

Hint: Given two distinct points $$(x_1,y_1),(x_2,y_2) \in [0,1]\times[0,1]$$, try to give an example of a function of the form $$F(x,y) = f(x)g(y)$$ for which $$F(x_1,y_1) \neq F(x_2,y_2).$$ You might find it helpful to separately handle the case where $$x_1 \neq x_2$$ and the case where $$y_1 \neq y_2$$. Note that our function $$F$$ is allowed to depend on the choice of points.