Schwartz writes in his book "Radon Measures on Arbitrary Topological Spaces", p. 110: "But even a finite union of polish spaces need not be polish". The same statement can be also found here.
How should this statement be interpreted? If I have a Polish space $X$ and a finite union of Polish subspaces $X_i \subseteq X$ then all of them are $G_\delta$-subsets and their (finite) union is again a $G_\delta$-subset of $X$ thus a Polish space.
Should this statement be interpreted that the given Polish spaces $X_i$ are not necessarily contained in a joint Polish space $X$? But what topology do we impose on the union? It may not be a disjoint union (which would be again Polish with the disjoint union topology).
Example of a non-metrizable space $S$ with two subspaces $S_1, S_2 $, each homeomorphic to the real line $R$, such that $S=S_1\cup S_2$ .$$\text {Let } S= (Q\times \{0\})\cup ((R\backslash Q)\times \{1,2\})$$ where $Q$ is the rationals. Let $T$ be the usual topology on $R$. $$\text {For } t\in T \text { let } t^*=((t\cap Q)\times \{0\})\cup ((t\backslash Q)\times \{1,2\})=(t\times \{0,1,2\})\cap S.$$ $$\text {Let } B=\{t^*\backslash u :t\in T\land ( u \text { is finite})\}.$$ I will leave it to you verify the following :(1) $B$ is a base for a topology $V$ on $S$. (2) $V$ is not a Hausdorff topology.(3)With the topology $V$ on $S$,the subspaces $$S_j=(Q\times \{0\})\cup ((R\backslash Q)\times \{j\}), \text { for } j\in \{1,2\}$$ are each homeomorphic to $R$ (by projection onto the first co-ordinate.)