I am try to answer the following question:
Let $E \subseteq \mathbb R$ and suppose for each $n \in \mathbb N$, there is a closed set $F_n\subseteq E \cap [-n,n]$. Show that $ F=\displaystyle\bigcup_{n=1}^\infty \left\lbrace x \in F_n : |x| \geq n-1 \right\rbrace$ is closed.
Here is my attempt:
Set $H_n = \left\lbrace x \in F_n : |x| \geq n-1 \right\rbrace$ and notice that $H_n \subseteq E \cap \big([-n,-n+1]\cup [n-1,n] \big)$.
Let $(x_n)_{n=1}^\infty \subseteq F$ with $x_n \to x \in \mathbb R$. Since $(x_n)_{n=1}^\infty$ converges in $\mathbb R$, then $(x_n)_{n=1}^\infty$ is a Cauchy sequence. Thus, we can choose an $N \in \mathbb N$, such that $|x_m - x_n |< 1$, whenever $m,n \in \mathbb N$ with $m,n > N$
Also, $x_n \in [-k, -k+1] \cup [k-1, k] $ for some $k \in \mathbb N$. If $x_n \in [-k, -k + 1] $ then the sequence eventually lies in the open interval $(-k-1,-k+1) $. Likewise, if $x_n \in [k-1,k]$, then the sequence eventually lies in the open interval $(k -1,k +1) $. In either case, all but finitely many term lies in $H_{k}\cup H_{k+1}$. $H_{k}\cup H_{k+1}$ is a closed set, hence the limit of the sequence must be one of the two and so the must be in $F$. Thus, $F$ is closed.
Obviously the endpoints $-k+1$ and $k-1$ are not included. So should take a $\varepsilon$ of 2 instead of 1? Also, any suggestions on making it lest clunky?