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

By Subsequences §subsequence, a subsequence of $(x_{n})$ is the composite of the sequence $(x_{n})$ with a strictly increasing map $\sigma:\mathbb{N}\to\mathbb{N}$; fix such a $\sigma$. The composite is a sequence in $X$ whose $k$-th term is $x_{\sigma(k)}$, by Basic Properties of Functions: Equality, Composition, Identity, Inverse and Restriction §composition; we write it $(x_{\sigma(k)})_{k\in\mathbb{N}}$.

Let $\varepsilon$ be a real number with $\varepsilon>0$. Since $(x_{n})$ converges to $x$, Convergent Sequence in a Metric Space §converges gives, for this $\varepsilon$, an $N\in\mathbb{N}$ with $d(x_{n},x)<\varepsilon$ for every $n\in\mathbb{N}$ with $n\ge N$. Let $k\in\mathbb{N}$ with $k\ge N$. By Monotone Sequences and Subsequences: Comparison of All Terms, Growth of the Indices, and Subsequences of Subsequences §index, $k\le\sigma(k)$, so $N\le\sigma(k)$ by the transitivity in Arithmetic and Order of the Natural Numbers §partial-order; applying the choice of $N$ with $n=\sigma(k)$ gives

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

Thus for every real number $\varepsilon>0$ there is $N\in\mathbb{N}$ with $d(x_{\sigma(k)},x)<\varepsilon$ for every $k\in\mathbb{N}$ with $k\ge N$, and by Convergent Sequence in a Metric Space §converges the subsequence $(x_{\sigma(k)})_{k\in\mathbb{N}}$ converges to $x$ in $(X,d)$. As $\sigma$ was an arbitrary strictly increasing map, this holds for every subsequence of $(x_{n})$.