Limit of a subsequence implies limit of a sequence

W.L.O.G let, $\{x_{n}\}_{n\in\mathbb{N}}$ be an increasing sequence.

To prove $\{x_{n}\}_{n\in\mathbb{N}}$ is convergent, it is enough to show $\{x_{n}\}_{n\in\mathbb{N}}$ is bounded above. (Why?)

Since, $\lim_{k \to \infty} x_{{n}_{k}}=a$ , one thing is clear , if $\{x_{n}\}_{n\in\mathbb{N}}$ converge, it must converge to $a$.

As we know a increasing sequence in $\Bbb{R}$ which is convergent, must converge to the sup of the set consisting the terms of the sequence.

With the above motivation, we claim that , $$x_n\le a$$ for all $n\in \Bbb{N}$

Prove by contradiction, suppose there is one $n_0 \in \mathbb{N} $ such that $x_n> a$

As $\{x_{n}\}_{n\in\mathbb{N}}$ is increasing sequence,

$$x_{m}> a $$ for all $m\ge n_0$

Now It's your task to contradict that $\lim_{k \to \infty} x_{{n}_{k}}=a$.