Let $(x_{n})_{n=m}^{\infty}$ be a sequence in $(X,d)$ which converges to some limit $x_{0}$. Then every subsequence $(x_{f(n)})_{n=m}^{\infty}$ of that sequence also converges to $x_{0}$.
My solution
Let $\varepsilon > 0$. Then there exists a natural number $N\geq m$ such that
\begin{align*}
n\geq N \Rightarrow d(x_{n},x_{0}) < \varepsilon
\end{align*}
Since $f:\textbf{N}\to\textbf{N}$ is stricly increasing, we conclude that $f(n) \geq n$.
Consequently, for the same $N\geq m$, the following relation holds
\begin{align*}
f(n) \geq n\geq N \Rightarrow d(x_{f(n)},x_{0}) < \varepsilon
\end{align*}
whence we conclude that $x_{f(n)}$ converges to $x_{0}$ as well.
I am mainly concerned with the wording of the proof. Can someone point out any flaw?
Best Answer
Your proof (and the wording as well) is fine. You proved that for all $\varepsilon > 0$, there exists an $N \in \mathbb{N}$ such that $n \geq N \implies x_{f(n)} \in B(x_0, \varepsilon)$, which shows exactly the convergence you desired.