[proofplan] We use the $\varepsilon$ definition of convergence in the metric space $(X,d)$. Given an arbitrary tolerance $\varepsilon>0$, convergence of the original sequence supplies an index $N$ after which every $x_n$ lies within distance $\varepsilon$ of $x$. A subsequence has strictly increasing natural-number indices, so its $k$th index satisfies $n_k\ge k$; hence all subsequence terms with $k\ge N$ come from the tail of the original sequence. This proves the same $\varepsilon$ condition for the subsequence. [/proofplan] [step:Use the tail estimate for the original convergent sequence] Let $\varepsilon>0$ be arbitrary. Since $x_n\to x$ in $(X,d)$, by the definition of convergence in a metric space there exists $N\in\mathbb{N}$ such that for every $n\in\mathbb{N}$ with $n\ge N$, \begin{align*} d(x_n,x)<\varepsilon. \end{align*} [/step] [step:Show the subsequence indices eventually lie in that tail] Let $(n_k)_{k=1}^\infty$ be a strictly increasing sequence of natural numbers. We claim that $n_k\ge k$ for every $k\in\mathbb{N}$. [guided] The only property of a subsequence that we need is the growth of its index sequence. Since $(n_k)_{k=1}^\infty$ is a strictly increasing sequence in $\mathbb{N}$, it cannot stay behind the identity sequence $k\mapsto k$. More precisely, we prove by induction that \begin{align*} n_k\ge k \end{align*} for every $k\in\mathbb{N}$. For $k=1$, the number $n_1$ is a natural number, and Androma's convention is that $\mathbb{N}$ starts at $1$. Hence $n_1\ge 1$. Now assume for some $k\in\mathbb{N}$ that $n_k\ge k$. Since the sequence $(n_k)_{k=1}^\infty$ is strictly increasing and its terms are integers, we have $n_{k+1}>n_k$, hence $n_{k+1}\ge n_k+1$. Combining this with the induction hypothesis gives \begin{align*} n_{k+1}\ge n_k+1\ge k+1. \end{align*} By induction, $n_k\ge k$ for every $k\in\mathbb{N}$. Therefore, if $k\ge N$, then \begin{align*} n_k\ge k\ge N. \end{align*} This is exactly what is needed: every sufficiently late subsequence term is also a sufficiently late term of the original sequence. [/guided] Indeed, since $n_1\in\mathbb{N}$, we have $n_1\ge 1$. If $n_k\ge k$, then strict increase and integrality give $n_{k+1}\ge n_k+1\ge k+1$. Hence $n_k\ge k$ for all $k\in\mathbb{N}$ by induction. In particular, for every $k\ge N$, \begin{align*} n_k\ge k\ge N. \end{align*} [/step] [step:Apply the original tail estimate to the subsequence] Set $K:=N$. If $k\in\mathbb{N}$ and $k\ge K$, then $n_k\ge N$ by the preceding step. Applying the tail estimate from the original sequence with $n=n_k$ gives \begin{align*} d(x_{n_k},x)<\varepsilon. \end{align*} Since $\varepsilon>0$ was arbitrary, this is precisely the definition of $x_{n_k}\to x$ in $(X,d)$. Therefore every subsequence of $(x_n)_{n=1}^\infty$ converges to $x$. [/step]