A sequence converges if and only if every subsequence converges?

Let $x_n \to x$. Then given $\varepsilon> 0$, there exists an $N \in \mathbb N$ such that $|x_n - x| < \varepsilon$ for all $n \geq N$. In words, it means that if we go out far enough, $N$ terms, we can talk about the rest of the terms of the sequence as being close enough, within $\varepsilon$, to the limit, $x$.

If you take any subsequence, say $(x_{n_k})_{k\in\mathbb N}$, then we can say that the $N^{th}$ term of the subsequence is at least, or beyond, the $N^{th}$ term of the actual sequence. Thus, it shares the same property that the terms of the sequence are within a desired distance from the limit of the main sequence.

Let $\epsilon > 0$ be arbitrary. Since $x_n \rightarrow x \text{ as } n \rightarrow \infty$, $\exists \; N \in \mathbb{N}$ such that $n \geq N \Rightarrow d(x_n, x) < \epsilon$.

Now, let $\{x_{n_k}\}_{k \in \mathbb{N}}$ be a sub-sequence of $\{x_n\}$. The part I'm writing next will give you intuition as to why the sub-sequence will also thin out to $x$ like the parent sequence.

$\exists \; k \in \mathbb{N}$ such that $n_k \geq N$. If not, then $N$ would be an upper bound for the $\textbf{strictly increasing}$ sequence of indices $\{n_1, n_2, \ldots\}$ and the sub-sequence won't have infinite terms. Try to understand this. This is the definition of a subsequence: that the indices are strictly increasing, i.e, $n_1 < n_2 <\ldots$.

Since $n_k \geq N$, we have that $d(x_{n_k}, x) < \epsilon$