Baire one extension of continuous functions I struggle with the following comment in Sierpinski's Hypothèse du continu, p. 49. 
For every continuous function $f(x):X \rightarrow \mathbb{R}$, where $X \subseteq \mathbb{R}$, there exist a Baire one function $g(x): \mathbb{R} \rightarrow \mathbb{R}$ such that $g(x) = f(x)$ for all $x \in X$.
What if both $X$ and $\mathbb{R}\backslash X$ are dense and uncountable ?

I think it can be proved by the following:
For arbitrary $X\subset\mathbb{R}$, continuous $f:X\longrightarrow\mathbb{R}$ can be extended to a function $F:\mathbb{R}\longrightarrow\mathbb{R}$ such that $F^{-1}(A)$ is a $G_\delta$ set in $\mathbb{R}$ for every closed $A\subset\mathbb{R}$ (a Lebesgue-one function such that $F|_X=f$).
Then the Lebesgue-Hausdorff theorem implies that the Lebesgue-one function $F$ is also Baire-one.
