Suppose that every subsequence of $X=x_{n}$ has a subsequence converge to $0$ show that $\lim X=0$ Please check my proof 
Suppose we have every subsequence $x_{n}$ converge to $0$
We have 
\begin{align}x_{1}:&\; |x_{1}-0|<\frac{\epsilon }{n} \\
x_{2}:&\; |x_{2}-0|<\frac{\epsilon }{n} \\
x_{3}:&\; |x_{3}-0|<\frac{\epsilon }{n} \\
&\vdots\\
x_{n}:&\; |x_{n}-0|<\frac{\epsilon }{n} \end{align}
then 
$|x_{1}-0|+|x_{2}-0|+|x_{3}-0|+....|x_{n}-0| <\frac{\epsilon }{n}+\frac{\epsilon }{n}....+\frac{\epsilon }{n}=\epsilon $
therefore $\lim X = 0$
 A: Your proof is incorrect. Saying that every subsequence has a subsequence converging to $0$ is not the same as every subsequence has every of its subsequences converging to $0$. But you appear to be using the former, leading to a flawed argument.
Here is an outline of a correct proof. Place your mouse over the hidden text areas to reveal their contents (i.e., a more detailed argument).
1). Show that $(x_n)_n$ is bounded.

 By contradiction: assume it is not. Then, for every $M>0$ and $N>0$ there exists $n\geq N$ such that $\lvert x_n\rvert \geq M$. In particular, this allows us to extract a subsequence $(x_{\varphi(n)})_n$ such that $\lim_{n\to\infty}\lvert x_{\varphi(n)}\rvert = \infty$. But this subsequence then does not have a subsequence converging to $0$, contradiction.

2). Use the fact that a bounded sequence converges if, and only if, it only has one limit point. To do that, take any limit point $\ell$ of $(x_n)_n$ (we know that there exists at least one, since $(x_n)_n$ is bounded), and show that necessarily $\ell=0$.

 Let $\ell$ be a limit point: by definition, there exists a subsequence of $(x_n)_n$ converging to $\ell$. Now this subsequence itself has, by hypothesis, a subsequence converging to $0$. But it also converges to $\ell$, since it's a subsequence of something converging to $\ell$: by uniqueness of the limit, $\ell=0$. Therefore, $0$ is the only limit point of $(x_n)_n$, meaning that $(x_n)_n$ converges to $0$.

