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.
Hint 1: Take a closed set in $C \subset \mathbb{P}^n$, and call $C_i=C \cap U_i$. Then, by definition, each $C_i$ is given as the zero set of some polynomials in $k[x_0/x_i, \ldots, x_n/x_i ]$ (where $x_i/x_i$ is omitted). Now, take those polynomials, and homogenize them: i.e., for a fixed polynomial $P \in k[x_0/x_i, \ldots, x_n/x_i ]$ you consider the biggest power $x_i^d$ appearing at the denominator, and then you consider $x_i^d P \in k[x_0,\ldots,x_d]$. It is a homogeneous polynomial of degree $d$.
Now, what can you do with all of these polynomials?
Hint 2: You can make it easier if you do a couple of reductions: consider just $C$ irreducible. If you can deal with it, then you just have to discuss what polynomials give $C \cup D$, once both $C$ and $D$ are given by polynomials. Second, using induction on the dimension (i.e. you show your statement is true for $\mathbb{P}^n$ inductively on $n$), you can assume that your irreducible $C$ is not contained in any of the complements of the $U_i$'s (if it was, then it would be contained in $\mathbb{P}^{n-1}$). With these two reductions, you should be able to use the polynomials as in Hint 1 coming just from one of the $C_i$'s.
Best Answer
That claim is incorrect: rather, it should be that any polynomial in $k[x_1,\cdots,x_n,y_1,\cdots,y_m]$ which can't be written as a product of polynomials in $k[x_1,\cdots,x_n]$ and $k[y_1,\cdots,y_m]$ defines a subvariety not closed in the product topology.
Your guess needs a little refinement in order to work: if you're working over a field $k$ and your product is really $\times_k$, this is true. Embed $X$ and $Y$ in to $\Bbb A^n_k$ and $\Bbb A^m_k$ respectively, which induces an embedding $X\times_k Y \to \Bbb A^n_k\times_k \Bbb A^n_k$. Pick some subvariety $Z$ of $\Bbb A^{n+m}_k$ which is not closed in the product topology and intersects $X\times_k Y$ with dimension at least 1. Then $Z\cap X\times_k Y$ cannot be a closed subvariety of the product topology on $X\times_Y K$.
(Note that this is false if we work over the integers and $X=Y=\operatorname{Spec}\Bbb Z$. The product here is over $\operatorname{Spec} \Bbb Z$ so $X\times Y = X =Y$, which is definitely affine, has the product topology, and is not finite. It's always good to specify your base!)