I have to show that, if ${ X_n }$ is a sequence of real numbers and $E$ denotes the set of all numbers $z$ that have the property that there exists a subsequence ${ X_{n_k} }$ convergent to $z$, then $E$ is closed.
So I have to show that any accumulation point of $E$, say $x$, is in the set $E$. By the definition of accumulation points of $E$ we have that for every $c>0$ the intersection $$(x-c,x+c) \cap E$$
contains infinitely many points. So there exists a subsequence ${ X_{n_k} }$ convergent to some $z$ arbitrary close to $x$. If ${ X_{n_k} }$ converges to $z$ then for all $\epsilon >0 $ there is an integer $N$ so that $$ \lvert { X_{n_k} } - z \rvert < \epsilon$$
whenever $n\geq N$. Choose $z$ such that $z>x>0$, then it follows that $$ \lvert { X_{n_k} } - x \rvert < \lvert { X_{n_k} } - z \rvert < \epsilon$$ whenever $n\geq N$. Thus $x$ converges to $x_{n_k}$ as well, so $x\in E$.
QED
Does my idea make sense? Is it rigorous? Any pointers/hints would be greatly appreciated.