Let $A: [a,b]\times[a,b] \to \mathbb{R}$ be a continuous function. Does it follow that, for each $y \in [a,b]$, the function
$$A_y: [a,b] \ni x \mapsto A(x,y) \in \mathbb{R}$$
is continuous (assuming Euclidean metric space)? I believe the answer is yes.
To establish continuity of $A_y$, fix $\epsilon > 0$ and $x \in [a,b]$. Now find (by continuity of $A$) a $\delta > 0$ so that $d((x,y), (x',y')) < \delta \implies \lvert A(x,y) - A(x',y') \rvert < \epsilon$.
Now, $|x-x'| < \delta \implies d((x,y), (x',y)) < \delta \implies \lvert A(x,y) - A(x',y) \rvert < \epsilon$, as desired.
Is this proof correct? Does it generalize to arbitrary topological spaces?
Your proof is correct. The result is true for general topological spaces too. If $A:X \times Y \to Z$ is continuous, where $X,Y,Z$ are topological spaces, then the map $X\ni x \mapsto A(x,y)$ is continuous for all $y\in Y$.