Prove $a_0\in A$ 
Let $\{ a_n \}_{n=1}^{\infty}$ be a sequence and $A=\{a|\exists a_{n_j}\rightarrow a\}.$ Suppose there exists $\{ x_n \}_{n=1}^{\infty}\subseteq A \ s.t.x_n \rightarrow a_0$. Prove $a_0\in A.$

I tried taking subsequence of $x_n$ which converges to $a_0$, and a subsequence of $a_n$ which converges to the subsequence of $x_n$ and do some algebraic manipulations, but got stuck.
Any help appreciated.
 A: Fix $\varepsilon>0$. 
Since $x_n\rightarrow a_0$, we have 
$$\text{For}\ \,\frac\varepsilon2>0,\ \exists N\in\mathbb N\ \text{such that}\ \forall n\geq N,\ \text{we have}\ |x_n-a_0|<\frac\varepsilon2$$
Because $x_n\in A$, there exists a sequence $\{a_{n_j}\}$ with  $a_{n_j}\rightarrow x_n$, and thus
$$\text{For}\ \,\frac\varepsilon2>0,\ \exists J\in\mathbb N\ \text{such that}\ \forall j\geq J,\ \text{we have}\ |a_{n_j}-x_n|<\frac\varepsilon2$$
Now by triangle inequality,
$$|a_{n_j}-a_0|\leq|a_{n_j}-x_n|+|x_n-a_0|<\frac\varepsilon2+\frac\varepsilon2=\varepsilon$$
and this shows that $a_{n_j}\rightarrow a_0$, hence $a_0\in A$

This is not a formal proof but an idea of how to use the triangle inequality.
A: Since you are not really saying where any of these elements live, I will assume a metric space, but you can do a fairly similar argument by showing $a_0$ is an accumulation point of $x_n$ which are accumulation points of $\{a_n \}$.
Note that for each $n$, there exists a subsequence $a_{n_j} \to x_n.$
Now, for $x_n$, consider the ball of radius $\frac{1}{n}$ centered at $x_n$, there exists $j_n$ so that for all $m \geq j_n$, $a_{n_m} \in B(x_n, \frac{1}{n}).$
Then consider the sequence $\{a_{j_n}\}_{n=1}^{\infty}.$ You can show that this sequence converges to $a_0$. For all $\epsilon > 0,$ there exist $n$ so that $d(x_n,a_0) > \frac{\epsilon}{2}$ (by convergence of $x_n$) and $d(x_n,a_{j_n}) > \frac{\epsilon}{2}$ (by construction of the sequence).
Hence you have a subsequence of $\{a_n\}$ converging to $a_0$. 
