Let $X$ be a subset of $\textbf{R}$. Then the following two statements are equivalent:
(a) $X$ is closed and bounded.
(b) Given any sequence $(a_{n})_{n=m}^{\infty}$ of real numbers which takes values in $X$, there exists a subsequence $(a_{n_{j}})_{j=0}^{\infty}$ of the original sequence, which converges to some number $L$ in $X$.
MY ATTEMPT
The implication $(a)\Rightarrow(b)$ results from the application of the Bolzano-Weierstrass theorem and the characterization of closed sets.
More precisely, if $a_{n}$ is a sequence whose terms are entirely contained in $X$ and it is bounded, then it admits a convergent subsequence. Since $X$ is closed, such subsequence must converge to an element $x\in X$.
The converse implication $(b)\Rightarrow (a)$ is the one which bothers me.
Could someone please provide a proof for such statement?
For every $a\in\overline{X}$, there exists a sequence $(a_n)_{n=0}^\infty$ in $X$ converging to $a$. Then every subsequence of $(a_n)_{n=0}^\infty$ also converges to $a$. By assumption, we have $a\in X$. Hence $\overline{X}=X$ and $X$ is closed.
On the other hand, suppose that $X$ is unbounded, say unbounded from above. Then we can find a divergent monotonically increasing sequence $(b_n)_{n=0}^\infty$ in $X$. Apparently, none of any subsequences of $(b_n)_{n=0}^\infty$ converges, giving a contradiction.