The problem is in the step where you assume
$\lim_{k\to\infty}f_k(x_k)=F(\lim_{k\to\infty}x_k)\;.$
You write "by continuity of $F$", but you don't have $F$ on the left-hand side, but $f_k$. Also, you can't simply replace this by an argument "by continuity of $f_k$" because it's a different $f_k$ for every $k$. Both the function and the argument vary with $k$ – which is what makes this slightly "tricky".
[Edit in response to the comment:]
Here's a diagram to visualize what's going on:
$\begin{matrix} f_j(x_k)&\rightarrow&f_j(x)\\ \downarrow&\searrow&\downarrow\\ F(x_k)&\rightarrow&F(x) \end{matrix} $
The vertical arrows hold because the $f_j$ converge pointwise to $F$. The horizontal arrows hold by continuity, namely of the $f_j$ for the upper one and of $F$ for the lower one. What you're trying to prove is the diagonal arrow in the middle; that is, you're setting $j=k$, which amounts to taking the diagonal of $f_j(x_k)$, and then letting that common index go to infinity in one go. That's not simply the composition of two of the other arrows; it's a different operation. If this isn't immediately clear, I suggest going through this diagram with the counterexample that Brian gave in mind; then you should be able to see what goes wrong with the diagonal arrow if you don't have uniform convergence.