Proving that the mothersequence converges to $x$ if any subsequence contains a subsequence which converges to $x$ [duplicate]

Dear reader of this post,

I am currently working on some problems about sequences and their subsequences. I proved a claim and because this prove involves some elementary concepts, I would like to ask three related questions. The claim is as follows: For any real number $x$ and $x_m \in \mathbb{R}^{\infty}$, show that $x_m \rightarrow x$ if every subsequence of $(x_m)$ has itself a subsequence that converges to $x$.

My prove is as follows: Take any subsequence $x_{m_{k}}$. As stated, I know that $\exists$ a (sub)subsequence $x_{m_{k_{i}}} \rightarrow x $. Let $\bar{x}_{m_{k_{i}}}$ be the non-convergent part of the subsequence. Redefine $x_{m_{k}}$ as $\tilde{x}_{m_{k_{i}}} = \left\lbrace \bar{x}_{m_{k_{1}}},\bar{x}_{m_{k_{2}}},\ldots,x_{m_{k_{1}}},x_{m_{k_{2}}},\ldots \right\rbrace $. Redefine the mother sequence $x_m$ furthermore as $\tilde{x}_m = \left\lbrace \tilde{x}_{m_{1}},\tilde{x}_{m_{2 }},\ldots \right\rbrace$. Finally, take any $\epsilon >0 $. I know that for any $ \tilde{x}_{m_{k}}$ $\exists M \in \mathbb{N}$ for which $\forall i > M$ $| \tilde{x}_{m_{k_{i}}} - x | < \epsilon $. Thus, $x_m$ is convergent.

After having stated my prove, I'd like to ask my three questions:

• I think the prove requires each subsequence to contain infinitely many elements. Is this correct and does every subsequence contain indeed infinitely many elements in general?
• Do I really need to redefine the sequence $x_m$ and its subsequences $x_{m_{k}}$ and is this legitimate?
• If these two questions are answered positively, is my prove correct?

Thank you very much for your support. I am looking forward for your replies.

Here is an alternative method to prove the desired claim.

Theorem: If every subsequence of $(x_n)$ has a subsequence which converges to $x$, then $(x_n)$ converges to $x$.

Proof: Suppose $x_n$ does not converge to $x$. Then there is $\varepsilon > 0$ such that $|x_n - x| \geq \varepsilon$ for infinitely many $n$. Therefore, there is a subsequence $(x_{n_k})$ with $|x_{n_k} - x| \geq \varepsilon$ for all $k \in \mathbb{N}$. This is a contradiction as $(x_{n_k})$ is a subsequence of $(x_n)$ which does not have a subsequence which converges to $x$.