If every subsequence is convergent, prove that the sequence is convergent 
If every subsequence of a given sequence of real numbers is convergent, prove that the sequence is convergent.

Help me please. I could not understand how to solve this question.
 A: Since the sequence $\,\{x_2,x_3,...,x_n,...\}\,$ converges, then also the whole sequence converges and, of course, to the very same limit.
A: Your question is trivial unless you change the "If" to "If and only if".  I'll prove the "if and any if" version.
(Trivial direction) Any sequence is a subsequence of itself, so if all subsequences of a given sequence converge, so does the original sequence.
(Nontrivial direction) Suppose $(x_n)$ converges to $L$.  Let $(x_{n_i})_{i \geq 1}$ be a subsequence of $(x_n)$. Let $\epsilon > 0$.  Since $(x_n)$ converges to $L$, there exists $N \equiv N(\epsilon) \in \mathbb{N}^+$ with the property that if $i \in \mathbb{N}^+$ with $i \geq N$, then $|x_i-L|<\epsilon$.  This is from the definition of convergence of a sequence.  Now $n_i \geq i$, so if $j\geq i$, then $n_j \geq n_i \geq i$, and $|x_{n_j}-L| < \epsilon$. So the subsequence  $(x_{n_i})_{i \geq 1}$ converges to $L$.
There is nothing special about $\mathbb{R}$ in the proof.  It would work the same in any metric space (I'm rusty when it comes to non-metric spaces).
A: Hint: Recall that a sequence $\{a_n\}$ of real numbers converges to a limit $L$ when the following statement is true: for any $\epsilon>0$, there is an integer $N$ such that, for all $n>N$, we have $|L-a_n|<\epsilon$.
Now use a proof by contradiction. If the original sequence didn't converge, then there is some $\epsilon>0$ such that, no matter what integer $N$ you pick, there is some element $a_n$ of the sequence, where $n>N$, such that $|L-a_n|\geq \epsilon$.
What did we just construct ... ?
