Proof of A Subsequence of a Convergent Sequence Has the Same Limit

Let $\varepsilon$ be a real number with $\varepsilon>0$. Since $(x_m)_{m\in\mathbb{N}}$ converges to $x$, there is $N\in\mathbb{N}$ such that

$$d(x_m,x)<\varepsilon\qquad\text{for every }m\in\mathbb{N}\text{ with }N\le m.$$

Let $k\in\mathbb{N}$ with $N\le k$. Because $(n_k)_{k\in\mathbb{N}}$ is strictly increasing, Strictly Increasing Sequences of Natural Numbers Dominate Their Index gives $k\le n_k$. The order $\le$ on $\mathbb{N}$ is transitive by Properties of the Order on the Natural Numbers, so $N\le k$ and $k\le n_k$ yield $N\le n_k$. Applying the displayed inequality with $m=n_k$ gives

$$d(x_{n_k},x)<\varepsilon .$$

Thus for every real $\varepsilon>0$ there is $N\in\mathbb{N}$ such that $d(x_{n_k},x)<\varepsilon$ for every $k\in\mathbb{N}$ with $N\le k$. By Convergent Sequence in a Metric Space the subsequence $(x_{n_k})_{k\in\mathbb{N}}$ converges to $x$ in $(X,d)$.