Let $(M,d)$ be a metric space and $F_1,F_2$ be two disjoint closed subspaces of $M$. If $(a_n)_{n \in \mathbb{N}}$ is a Cauchy sequence such that $a_n \in F_1 \cup F_2$ for every $n$, and there is an infinite number of terms of the sequence in $F_1$, does it follow that there is only a finite number of terms of the sequence in $F_2$?
I thought about this question while doing some exercises about metric spaces. Here is what I've got so far. If $F_1$ and $F_2$ are complete metric spaces (with respect to the induced metric), then the claim seems to be true, because the existence of infinite terms of the sequence in $F_1$ implies the existence of a subsequence $(a_{n_i})_{n_i \in \mathbb{N}}$ of the original sequence which is contained in $F_1$, and since this subsequence is also Cauchy and $F_1$ is complete, it converges to some point in $F_1$, but from the fact that the original sequence was Cauchy, we conclude that the whole sequence converges to a point in $F_1$. Now, if there was an infinite number of terms in $F_2$, we would conclude by the same reasoning that $(a_n)$ would converge to some point in $F_2$, but the uniqueness of the limit would imply that such limit would be in $F_1 \cap F_2$, a contradiction.
If the above conclusion is correct, we know that the claim is also true if $M$ is complete, for this implies that both $F_1$ and $F_2$ are also complete.
So, if there is a counterexample, it must be in a non-complete metric space, but I haven't been able to come up with one so far. It seems that such a counterexample would require two closed sets having arbitrarily close elements, but with none in common. Maybe the claim is trvially false, but i couldn't prove it or disprove it in general.
Thanks in advance for the help and attention.
In $(0,1]$, set $x_n:=1/n$ for each $n$, $F_1:=\{1/n \mid \text{$n$ is even}\}$, and $F_2:=\{1/n\mid\text{$n$ is odd}\}$.
It's clear that $(x_n)$ is a Cauchy sequence and that $F_1$ and $F_2$ are disjoint subsets of $(0,1]$ such $x_n\in F_1\cup F_2$ for each $n$ and the sets $\{x_n\}\cap F_1$ and $\{x_n\}\cap F_2$ are infinite. Can you show that $F_1$ and $F_2$ are closed in $(0,1]$?