Show that every Cauchy sequence can have at most one limit point. Show that every Cauchy sequence can have at most one limit point.

My solution

If a Cauchy sequence $x_{n}$ admits a limit point $L$, then there is a subsequence of $x_{n}$ which converges to $L$. Therefore $x_{n}$ converges to $L$. Since the limit of a sequence is unique in a metric space, the result holds.

Can someone check if is it right? An answer based on $\varepsilon-\delta$ definitions would be nice.

Edit 1

My answer is based on the following result:


  Let $(X,d)$ be a metric space and $x_{n}$ is a Cauchy sequence in it. Then, if $x_{n}$ admits a subsequence which converges to $x_{0}$, then the Cauchy sequence is also convergent and $x_{n}\rightarrow x_{0}$.


Edit 2

Since my question has caused some confusion, I am going to explicit what I was looking for.

The proposition "if $L$ is a limit point of the sequence $x_{n}$, then there is a subsequence of $x_{n}$ which converges to $L$" has been proved earlier as well as the other two results involved in the solution.

Maybe I should have asked for an alternative solution involving $\varepsilon-\delta$, but as it has been pointed out it does not make sense since they are already involved in the assumptions used to solve the problem.

Anyway, thank you guys for the feedback.
Assume there are two subsequence $ x_{\phi(n)} $ and $ w_{\psi(n)} $ such that

$$x_{\phi(n)}\to L_1$$
and
$$x_{\psi(n)}\to L_2$$

Given $ \epsilon&gt;0$,

we know that
$$\phi(n)\ge n ; \psi(n)\ge n$$

then there exists $N_1, N_2,N_3 \ge 0$ such that
$$n\ge \max(N_1,N_2,N_3) \implies$$

$$|L_1-L_2|=|L_1-x_{\phi(n)} +x_{\phi(n)}-x_{\psi(n)}+x_{\psi(n)}-L_2|\le \frac \epsilon 3+\frac \epsilon 3 +\frac \epsilon 3$$

We conclude that $$L_1=L_2$$
A Cauchy sequence is a convergent sequence so it has exactly 1 limit point, not at most 1.

Edit: If the sequence comes from real numbers.