You could take the sequence $a_n = (-1)^n$. What are the convergent subsequences?
It is necessary that a convergent sequence be a Cauchy sequence. Conversely, a Cauchy sequence with a converging subsequence converges.
Proof by contradiction
Suppose that $\{X_n\}$ does not converge to $\ell$. Then, there is $\varepsilon_0>0$ such that $$\forall N\in\mathbb N,\exists n=n(N) : n>N~~~and ~~~ |X_n -\ell|>\varepsilon_0 $$
For $N_1=1$ there exists $n_1$ such that
$$n_1>N_1 ~~~and ~~~ |X_{n_1} -\ell|>\varepsilon_0 $$
Taking successively $N_{k+1}> \max\{N_k, n_k,k+1\}$ there exists $n_{k+1}>N_{k+1}$ such that,
$$ |X_{ n_{k+1}} -\ell|>\varepsilon_0 $$
It is easy to see that, $\{X_{ n_k}\}_k$ is a subsequence of $\{X_{ n}\}_n$
since
$$ n_k< n_{k+1} \quad i.e ~~\text{the map }~~k\mapsto n_k~~~\text{Is one-to-one}$$
However, $$\forall k,~~ |X_{ n_{k}} -\ell|>\varepsilon_0 \qquad \text{and}~~~\{X_{ n_{k}} \}~~~\text{is bounded} $$
Therefore By Bolzano-Weierstrass Theorem's there exists $\{X_{ n_{k_p} }\}_p$ subsequence of $\{X_{ n_{k} }\}_k$ which converges to some limit $\ell_1 $
but $\{X_{ n_{k_p} }\}_p\to \ell_1$ is also a converging subsequence of $\{X_n\}_n$
By assumption, $\ell=\ell_1$ that is together with the fact $\{X_{ n_{k_p} }\}_p$ is a subsequence of $\{X_{ n_{k} }\}_k$ we have
$$0=\lim_{p\to\infty } |X_{ n_{k_p} }-\ell|>\varepsilon_0>0~~~\text{which is a CONTRADICTION}$$
Note that
$$\forall p,~~|X_{ n_{k_p} }-\ell|>\varepsilon_0$$
Since
$$\forall k,~~|X_{ n_{k}} -\ell|>\varepsilon_0$$
Best Answer
Here is an outline of a proof for you to fill out:
Theorem: Let $(x_n)$ be a sequence of real numbers. Let $x$ be a real number. Then $(x_n)$ converges to $x$ if and only if both of the following hold:
Lemma 1: If $(x_n)$ is a convergent sequence of real numbers, then $(x_n)$ is bounded.
Proof: Suppose $(x_n)$ converges to $x$. Then for any $\epsilon>0$ there is an $N>0$ such that …. In particular, letting $\epsilon = 1$ ….
Lemma 2: If $(x_n)$ is a sequence of real numbers converging to $x$, and $(x_{n_k})$ is any subsequence thereof, then $(x_{n_k})$ converges to $x$.
Lemma 3: If $(x_n)$ is a bounded sequence of real numbers, then $(x_n)$ has a convergent subsequence.
Lemma 4: If $(x_n)$ is a bounded sequence of real numbers that does not converge to $x$ then it must have a subsequence that converges in $(x,\to)$ or in $(\gets,x)$.