Lemma:
Let $f:(X,\mathcal{T})\longrightarrow(X',\mathcal{T'})$ be a mapping between topological spaces where $$X=\cup_{j=1}^m F_j$$ $ \text{ with }F_j\text{ closed for }\mathcal{T},\, f_{|F_j} \text{ continuous },\forall j=1,\ldots,m.$
Then $f$ is continuous.
I'd like to find a non-example to that when $X$ is a countably infinite union of closed sets.
My course already provides one that I find counterintuitive as I've mostly worked so far on metric spaces (the identity between $\mathbb{N}=\cup\{n\}$ with the cofinite topology and $\mathbb{N}$ with the discrete topology).
Is it possible to find one where both spaces are $\mathbb{R}$ with the standard topology ?
I've considered disjoint closed $X_n$ that increasingly cover $\mathbb{R}$ but in vain.
Take any function $f$ from $\mathbb R$ into itself. For each $x\in\mathbb R$, $f|_{\{x\}}$ is continuous and $\{x\}$ is closed. But $f$ doesn't have to be continuous.
As another exemple, consider $f\colon[0,1]\longrightarrow\mathbb R$ such that $f(0)=0$ and that $f(x)=1$ otherwise. Then $f$ is discontinuous, but if you define $I_0=\{0\}$ and, for each natural $n$, $I_n=\left[\frac1{n+1},\frac1n\right]$, then the restriction of $f$ to each $I_n$ is continuous, whereas $f$ is discontinuous.