The statement of Baire grand theorem gives a characterization of Baire class 1 functions between a completely metrizable separable space (aka Polish space) and a separable metrizable space. The statement is the following:

Let $X$ be a Polish space, $Y$ separable metric space and $f:X\rightarrow Y$. Then the following are equivalent:

1. $f$ is Baire class $1$, i.e. the preimages of open sets are $F_\sigma$ sets.
2. $f\upharpoonright F$ (the restriction of $f$ on $F$) has a point of continuity for every non-empty closed set $F\subseteq X$.
3. $\inf\{\text{osc}_{f\upharpoonright F}(x) \mid x \in F\} = 0$ for every non-empty closed set $F\subseteq X$.

I recall that $\text{osc}_f(x) = \inf\{\text{diam}(f(U))\mid x \in \text{dom(f)}, U \text{ open nbhd of }x\}$.
The equivalence $2 \Leftrightarrow 3$ is not that interesting and is a straightforward application of Baire category theorem (which guarantees us that every Polish space is hereditary Baire). Also $1\Leftrightarrow 2$ relies on Baire category theorem.
But if we drop the assumption that $X$ is completely metrizable then the equivalence of the statement above no longer holds. Fix for example a bijection $\varphi:\mathbb{Q}\rightarrow \mathbb{N}$, this function is nowhere continuous and its oscillation is infinite in every point, but it's still Baire class $1$, as every function with domain $\mathbb{Q}$ is Baire class 1.

Now I was wondering if we could get also a counterexample for the directions $2\Rightarrow 1$ and $3\Rightarrow 1$. Are there separable metrizable spaces $X,Y$ and a function $f:X\rightarrow Y$ such that $f\upharpoonright F$ has a point of continuity (resp. has arbitrarily small oscillations) for every closed $F\subseteq X$ but $f$ is not Baire class 1?
Do we need to assume some form of Choice to produce such counterexamples? Any ideas, suggetions?

Thanks!