Continuous functions between metric spaces are equal if they are equal on a dense subset
This is correct. Suppose $f$ and $g$ are continuous functions on a metric space $X$ and agree on a dense subset $Y$. For any $x\in X$, we have some sequence $(y_n)$ in $Y$ such that $y_n\to x$, so $f(y_n)\to f(x)$ and $g(y_n)\to g(x)$. Since $f(y_n)=g(y_n)$ for all $n$, this implies $f(x)=g(x)$.
Yes. The set of points where the functions $f,g\colon X\to Y$ agree is closed and contains a dense subset. The closure of a dense subset is $X$. (This does not require metric spaces, it is sufficient that $X$ is any topological space and $Y$ is Hausdorff)