"Let $f, g: \mathbb{R} \longrightarrow \mathbb{R}$ continuous functions such that $f(r) = g(r)$ for all $r \in \mathbb{Q}$. Proof that $f(x) = g(x)$ for all $x \in \mathbb{R}$".
My resolution: $\mathbb{Q}$ in dense in $\mathbb{R}$, so $\mathbb{R} \subset \bar{\mathbb{Q}}$. Let $x \in \mathbb{R}-\mathbb{Q}$, exist a sequence $(x_{n})$ with $x_{n} \in \mathbb{Q}$ for all $n \in \mathbb{N}$ such that $\lim x_{n} = x$. Since $f, g$ are continuous, $\lim f(x_{n}) = f(x)$ and $\lim g(x_{n}) = g(x).$ But $f(x_{n}) = g(x_{n})$ for all $n \in \mathbb{N}$, so $f(x) = g(x)$ for all $x \in \mathbb{R}-\mathbb{Q}$. The proof is complete.
Is this correct?
Yes, it's correct. Alternative method:
By considering $h = f - g$ it's enough to show that "if $h$ is continuous, and has value equal to $0$ on a dense set $\mathbb{Q}$, then it has value equal to $0$ everywhere".
Consider $h^{-1}(\{\mathbb{R} \setminus \{0\}\})$. Since the continuous preimage of an open set is open, it must be the case that this set is open. But it certainly doesn't contain any members of $\mathbb{Q}$, because $h$ would be $0$ on any of those members; so it is an open set which does not intersect a certain dense set, so it is empty.
This proof didn't require any use of the metric-space properties of $\mathbb{R}$; it's purely topological.