Using the least upper bound property of real numbers prove that every closed interval $[a,b]\space (a\lt b)$ is compact.
I tried solving this problem like this:
We know that $K\subset \Bbb R$ is compact iff for every sequence in $K$ there exists a convergent subsequence such that it converges, at a point in $K.$ Now, with this definition of compact set in mind, we try to prove the required claim. Let us consider an arbitrary closed interval $[a,b]$ with $a\lt b.$
Now, if $K=[a,b]$, we note that $\forall k\in K,$ $a\leq k\leq b.$ This means, that $K$ is bounded.
Now, if say, $(x_n)$ is a convergent sequence in $K$, it means that $(x_n)$ has a convergent subsequence in $K$ by Bolzano-Weirstrass's Theorem. Let this convergent subsequence be $(x_{n_k}).$ Now, we observe that, $a\leq x_{n_k}\leq b.$ In other words the term of this convergent subsequence is bounded as well. Now, if $\lim x_{n_k}=L,$ we have that $a\leq L\leq b$, by Limit -Ordering Property. As, $(x_n)$ is an arbitrary sequence in $K=[a,b]$ we have that the definition of a compact set is satisfied by $K.$ So, $[a,b]$ is a compact set.
I hope my solution is correct. But the problem, is, in my proof, never have I used the lub property of real numbers directly. It might have been used in some indirect way, because, if I'm not mistaken the least upper bound property of the real numbers is just a different way of stating, the Axiom of Completeness (AoC) and since AoC is such a fundamental result, that it might be justified in saying, most of the basic theorems of sequences and series in real analysis, rests upon this so-called axiom. So, can I consider my proof to be apt as an answer to this question.
Also, I want to know if my proof can be improved further or not. Lastly, if I have made any errors in my solution please do let me know.
Your proof is correct and you have used LUB implicitly by using Bolzano-Weierstrass which requires LUB or some equivalent statement in its proof.
If this is an exercise for course work however, your proof might not be what the teacher has in mind. In my experience those types of exercises require you to directly use the property and (ideally) only that property. If you suspect this is the case, you may try a proof along the following lines:
Let $x_n$ be a sequence in $K = [a,b]$ and consider the set $X = \{ y \in K : y \leq x_n \text{ for infinitely many } n\}$ which has a least upper bound $\overline x$ in $K$. Can you now construct a convergent subsequence of $x_n$ with limit $\overline x$?