Showing a set is dense in $C([0,1]\times [0,1])$ 
Let $A = \{h\in C([0,1]\times [0,1]) :\\h(x,y) = \sum_{k=1}^n f_k(x)g_k(y), n\ge1, f_k,g_k\in C([0,1])\},$ where $(C([0,1]),d_\infty)$ is the metric space of continuous functions on $[0,1]$ equipped with the supremum metric. Show that $A$ is dense in $C([0,1]\times [0,1])$ under the supremum metric.

I know the Stone-Weierstrass theorem. However, I'm not sure if it's useful here. Let $f \in C([0,1]\times [0,1])$ and $\epsilon > 0.$ Since $[0,1]\times [0,1]$ is compact, $f$ is uniformly continuous, so $\exists \delta > 0$ so that for all $(x,y), (x',y')\in [0,1]\times [0,1], \lVert(x,y)-(x',y')\rVert_2<\delta$ implies $| f(x,y)-f(x',y')| < \epsilon.$ I want to find a function $g \in A$ so that $\lVert f-g\rVert_\infty < \epsilon.$ However, I'm not sure which functions to choose and what value of $n$ to choose to define the function $g$.

To use SW we need to verify that i) $A$ is an algebra of continuous functions on $K=[0,1]\times [0,1],$ ii) $A$ separates points on $K,$ and iii) $A$ vanishes nowhere on $A.$
i) is elementary; I'll leave it to you. ii) Assume $(x_1,y_1), (x_2,y_2)$ are distinct points in $K.$ Then either $x_1\ne x_2$ or $y_1\ne y_2.$ Suppose the former. Let $h(x,y)=x\cdot 1.$ Then $h\in A$ and $h(x_1,y_1) = x_1\ne x_2=h(x_2,y_2).$ Similarly for $y_1\ne y_2.$ iii) Just take $h\equiv 1$ on $K.$
SW then says $A$ is dense in $C(K).$
