In a metrizable space, the union of a chain of closed subsets of is F$_\sigma$ 
Let $(X,\tau)$ a metrizable topological space and $\mathcal{F}$ a collection of closed subsets of X which is linearly ordered by inclusion. Show that $\bigcup \mathcal{F}$ is F$_\sigma$.

Here is my attempt to prove this: 
Let $d$ a metric that generates $\tau$. Suppose that is not possible to write $\bigcup\mathcal{F}$ as a countable union of closed sets, lets say $\bigcup F_n$, $F_n \in \mathcal{F}\enspace \forall n$.
Let $F\in\mathcal{F}$. If $K \in \mathcal{F}$, is such that $F\subset K$, there is $x \in K$ such that $d(x,F) >0$. So, it would be possible to obtain a non-countable set $\{x_K : K \in \mathcal{F} \}$ such that $d(x_K, F)$ is monotone increasing in $\mathbb{R}$. Then we obtain a non-countable collection of of disjoint open intervals in $\mathbb{R}$. 
I am not sure if everything so far is right, and I don't have any other ideas to prove the assertion. Can anyone help?
 A: Nate's observation is essentially this:
Let $X$ be a countably tight space, i.e. for every subset $A \subseteq X$
$$\forall x \in \overline{A} :\exists A' \subseteq A: (|A'| \le \aleph_0) \land (x \in \overline{A'})$$
Note that all metric spaces satisfy this, we can even take a convergent sequence from $A$ converging to $x$, using a countable local base.
Suppose $\mathscr{F}$ is a family of closed sets, linearly ordered by inclusion.
Suppose that $\mathscr{F}$ has an up-cofinal countable subset:
$$\exists \mathscr{F}' \subseteq \mathscr{F}: (\left|\mathscr{F}'\right| \le \aleph_0) \land (\forall F \in \mathscr{F} \exists F' \in \mathscr{F}': F \subseteq F')$$ 
In that case $\bigcup \mathscr{F} = \bigcup \mathscr{F'}$ (right to left inclusion is immediate and left to right inclusion folllows from cofinality) and the last set is an $F_\sigma$ by definition.
If there is no countable cofinal subset: $F= \cup \mathscr{F}$ is closed. 
Let $x \in \overline{F}$, then the countable tightness guarantees that we have a countable subset $C$ of $F$ such that $x \in \overline{C}$. For every $c \in C$ pick $F_c \in \mathscr{F}$ with $c \in F_c$. Then $\mathscr{F}' = \{F_c:  \in C\}$ is countable so cannot be cofinal, which means that there is a $G \in \mathscr{F}$ such that for all $c$, $G \nsubseteq F_c$ and, as we have a linearly ordered collection, for this $G$ we have $F_c \subseteq G$.
Now: $$x \in \overline{C} \subseteq \overline{\bigcup\{ F_c: c \in C\} } \subseteq \overline{G} = G$$ 
as $G \in \mathscr{F}$ is closed. So $x \in F$ and so $F = \cup \mathscr{F}$ is closed ( so also an $F_\sigma$).
A: I don't think your proof works.  You can't ensure that, for $K \subset K'$, we have $d(x_K, F) \le d(x_{K'}, F)$.  Consider the chain $\{0\} < \{0,1\} < \{0, 1, 1/2\}$.   Also, even if you do get a monotone increasing chain in $\mathbb{R}$, that's no contradiction ($\mathbb{R}$ itself is such a chain) and it doesn't imply the existence of uncountably many disjoint intervals.
Here's a hint, though.  Set $A = \bigcup \mathcal{F}$.  If there is a sequence $F_1 \subset F_2 \subset \dots$ which is cofinal, i.e. such that for every $F \in \mathcal{F}$ there is $n$ such that $F \subseteq F_n$, then we are done because $A = \bigcup_n F_n$.  So suppose not.  In this case you can show that $A$ is closed.  For let $x$ be a limit point of $A$.  You can then find a sequence $x_n \in A$ converging to $x$.  But each $x_n$ is contained in some $F_n \in \mathcal{F}$; what can you do with the sequence $F_n$?
