I want to be sure of my reasoning. Can someone please give feedback on my proof?
Let $N$ be a sequence of non-negative integers s.t. no integer occurs more than a finite nr. of times; that is, $$\forall m \quad \{i:N_i=m\}\; \text{is finite.}$$Then if $S_n$ is any sequence, $S_{N_i}$ is a subsequence. If $\{S_n, n\in \Bbb N\}$ is a sequence in a topological space, and $N$ is an arbitrary sequence of $\Bbb N$, then $\{S_{N_i}, i\in \Bbb N\}$ is either a subsequence of $\{S_n, n\in \Bbb N\}$ or else has a cluster point.
Proof
Let $N$ arbitrary sequence with described property and $\{S_n, n\in \Bbb N\}$ a sequence in topological space.
We need to show that $\{S_{N_i}, i\in \Bbb N\}$ is either a subsequence of $\{S_n, n\in \Bbb N\}$ or else has a cluster point.
Suppose that $S_{N_i}$ is neither a subsequence of $S_n$ and has no cluster point.
If $S_{N_i}$ is not a subsequence of $S_n$, then there is $m\in \Bbb N$ and $x\in X$ s.t.
$S_{N_m} = x$ and $S_k\neq x$ for any $k\in \Bbb N$.
But this contradicts the choice of $i\in \Bbb N$ for each $m \in \Bbb N$ s.t. $N_i=m$. For surely if $N_i=m$, then $S_{N_i}=S_m$.
Hence $S_{N_i}$ is a subsequence of $S_n$ and it contradicts initial assumption. Therefore $S_{N_i}$ is either a subsequence or a cluster point.