In general, it's definitely not true that the countable union of nested closed subsets of a complete metric space is complete: the union of $[1/n, 1-1/n]$ for all $n\in \mathbb{N}$ is $(0,1)$, the union of $Q_n := \left\{\frac{r}{s}\,|\, (r,s) = 1,\, n! | s\right\}$ for all $n\in\mathbb{N}$ is $\mathbb{Q}$ etc. However, there is a situation I'm interested in where this is the case:
Suppose $F_0, F_1, F_2,\ldots $ is a sequence of closed subsets of some complete metric space $(X,d)$ with $F_i \subseteq F_{i+1}$ and also $\text{diam}(F_i)\to \infty$ as $i\to\infty$. Then the union of the $F_i$s is closed, since any convergent sequence $(x_i)$ inside their union is bounded and so there is some $k$ such that $x_j \in F_k$ for all $j \in \mathbb{N}$ and since $F_k$ is closed, the limit $x$ of the sequence must lie inside $F_k$ and thus their union.
My question is: is the corresponding statement true for complete metric spaces? This might be really obvious but I can't tell if what I said above can be converted into a proof of the following statement:
Suppose $X_0, X_1, X_2, \ldots $ is a sequence of nested complete metric spaces, in the sense that there is a distance preserving function $\sigma_{i,j}:X_i\to X_j$ for any $i<j$ (and we can assume WLOG that the underlying sets have $X_i\subseteq X_j$), such that $\text{diam}(X_i)\to \infty$ as $i\to \infty$. Then the "union space", with underlying set $\bigcup_{i\geq 1} X_i$ and metric inherited from each $X_i$, is complete.