Looks like the correct approach. Note that what you're showing is that
$$V \times W = \{(x_1, \dots,x_n,y_1, \dots, y_m) \in k^{n+m} \mid f_1(x) = \dots = f_s(x) = 0, g_1(y) = \dots = g_t(y) = 0 \}.$$
If you feel uncomfortable because the defining equations should be polynomials in $x, y$ (instead of just in $x$ for the $f$'s and just in $y$ for the $g$'s), you might want to make things explicit.
Say $\hat {f_i}(x_1,\dots,x_n,y_1,\dots,y_m) = f_i(x_1,\dots,x_n)$ (i.e., $f_i$ explicitly considered as a polynomial in $x, y$) and similarly $\hat{g_j}(x_1,\dots,x_n,y_1,\dots,y_m) = g_j(y_1,\dots,y_m)$. Phrased this way, you're showing that
$$V \times W = \{(x_1, \dots,x_n,y_1, \dots, y_m) \in k^{n+m} \mid \hat{f_1}(x,y) = \dots = \hat{f_s}(x,y) = 0, \hat{g_1}(x,y) = \dots = \hat{g_t}(x, y) = 0 \},$$
but the right-hand side of this is of course still equal to
$$\{(x_1, \dots,x_n,y_1, \dots, y_m) \in k^{n+m} \mid f_1(x) = \dots = f_s(x) = 0, g_1(y) = \dots = g_t(y) = 0 \},$$
so that is really just a notational issue.
Here is just an outline.
Consider the ring extension $k \to A(X)$. Use Noether normalization to see $A(X)$ is integral over a subring isomorphic to $k[x_1, \dots, x_s]$, where $s$ is the dimension of $X$ (using the fact that definition of dim $X$ in terms of chains of irreducible closed subsets is the same as the Krull dimension of the coordinate ring $A(X)$). Do the same for $k \to A(Y)$ to produce another subring $k[y_1, \dots, y_t]$, where $t=dim Y$.
Next show that coordinate ring of $X \times Y$ is integral over a subring isomorphic to $k[x_1, \dots, x_s, y_1, \dots, y_t]$. Finally use the lying over and going up theorems to conclude that $A(X \times Y)$ also has dimension $s+t$.
Best Answer
Since $A(X)$ is finitely generated and integral over $k[x_1,\cdots,x_n]$, it is finite over this ring, and we may pick some finite collection of generators of this ring extension $a_1,\cdots,a_r$, each of which is algebraic over $k[x_1,\cdots,x_n]$. Then $K(X)=k(x_1,\cdots,x_n,a_1,\cdots,a_r)$. We make a similar construction for $Y$, where $b_1,\cdots,b_s$ are the generators for $A(Y)$ over $k[y_1,\cdots,y_m]$, and note that $K(Y)=k(y_1,\cdots,y_m,b_1,\cdots,b_s)$. Since $A(X\times Y)\cong A(X)\otimes A(Y)$, the elements $x_i,y_j,a_k,b_l$ generate $A(X\times Y)$ and we see that $$K(X\times Y)=k(x_1,\cdots,x_n,y_1,\cdots,y_m,a_1,\cdots,a_r,b_1,\cdots,b_s)$$ which is the same as saying that $$K(X\times Y)=k(x_1,\cdots,x_n,y_1,\cdots,y_m)(a_1,\cdots,a_r,b_1,\cdots,b_s).$$ This extension is algebraic since it is generated by algebraic elements, so the $x_i$ and $y_j$ considered together form a transcendence basis.