I have a metric space, and two closed subsets that are both completely metrisable (not necessarily for the same metric, that wouldn't be fun). I'm trying to prove that their union is still completely metrisable, but everything I try to do fails, to the point where I'm thinking that it might not be true.
Are there known counter examples of this assertion, or is there a way to prove it that I haven't thought of?
You can assume without loss of generality that the ambient space is complete, we don't need these sets to be closed. By a result of Alexandroff, a subset of a complete metric space is completely metrizable if and only if it is a $G_\delta$, a countable intersection of open sets.
If $F_1$ and $F_2$ are completely metrizable, there must be decreasing sequences of open sets $(U_n)$ and $(V_n)$ such that $F_1=\bigcap_n U_n$ and $F_2=\bigcap_n V_n$. But then $(F_1\cup F_2)=\bigcap_n (U_n\cup V_n)$, and the union is again completely metrizable by the result of Alexandroff.