As the interval is closed every sequence in the interval converges to some point $x$ in the interval, and every convergent sequence is a Cauchy sequence, hence $[x, y]$ is complete.
Is that correct? I thought so earlier but now I am not sure where I saw it mentioned that a closed interval/set implies that any sequence in that set converges in that interval/set.
It’s not true that every sequence in $[a,b]$ converges to a point of $[a,b]$: some sequences in $[a,b]$ don’t converges at all! For instance, take $[a,b]=[-1,1]$, and let $x_k=(-1)^k$. What is true is that if $\langle x_k:k\in\Bbb N\rangle$ is a convergent sequence in $[a,b]$, then its limit is also in $[a,b]$; proving this is a good elementary exercise.
There are many ways to show that $[a,b]$ is complete if $a<b$, depending on what tools you have available. For instance, it’s a theorem that every compact metric space is complete, and $[a,b]$ is certainly a compact metric space. Another theorem says that a closed subspace of a complete metric space is complete; if you know that theorem and know that $\Bbb R$ is complete in the usual metrix, it follows immediately that $[a,b]$ is complete.
For a direct proof, let $\langle x_k:k\in\Bbb N\rangle$ be a Cauchy sequence in $[a,b]$. Since $[a,b]$ is compact, $\langle x_k:k\in\Bbb N\rangle$ has a convergent subsequence, say with limit $y$. By the exercise that I mentioned above, $y\in[a,b]$. Now use the fact that $\langle x_k:k\in\Bbb N\rangle$ is Cauchy to prove that it must also converge to $y$, and you’ll have shown that every Cauchy sequence in $[a,b]$ converges in $[a,b]$.
