Maximally continuous extension of continuous functions from $\mathbb Q$ to $\mathbb R$ Let $f: \mathbb Q \to \mathbb R$ be a continuous function.
An extension of $f$ is a function $\tilde f: \mathbb R \to \mathbb R$ such that $\tilde f = f$ on $\mathbb Q$.
We say an extension $\tilde f$ of $f$ is maximally continuous if for any other extension $g$ of $f$, we have that if $g$ is continuous at $x \in \mathbb R$, then so is $\tilde f$.

Question: For any continuous function $f: \mathbb Q \to \mathbb R$, does there exist a maximally continuous extension of $f$?

Remark: One can always obtain an extension that is continuous at every point in $\mathbb Q$, see for example, the answer in this post by Fedor Petrov.
 A: There even exists a largest set $X$ to which $f$ can be continuously extended.
The trick is the following result (which I state here in more generality, to point out which topological assumptions one needs):
Theorem. Let $X,Y$ be topological spaces, where $Y$ is $T_3$, and let $D \subseteq X$ be dense. Let $f: D \to Y$ be continuous and assume that the following property is satisfied for each point $x \in X \setminus D$ (equivalently, each point $x \in X$):
$(*)$ There exists $y_x \in Y$ such that, for each net $(x_j)$ in $D$ that converges to $x$, the net $(f(x_j))$ converges to $y_j$.
Then $f$ has a (obviously, unique) continuous extension to $X$.
(I gave a proof of this in a topology course a year ago, but unfortunately I don't know a reference - maybe somebody else can help out with a reference?)
Application to your situation. For $D = \mathbb{Q}$, let $X$ be the set of all $x \in \mathbb{R}$ which satisfy property $(*)$. Then $f$ can be continuously extended to $X$, and this extension is clearly the largest one.
