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.

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).
