A direct proof is normally easiest when you have some obvious mechanism to go from a given hypothesis to a desired conclusion. (E.g. consider the direct proof that the sum of two convergent sequences is convergent.)
However, in the statement at hand, there is no obvious mechanism to deduce
that the sequence converges to $a$. This already suggests that it might be
worth considering a more roundabout argument, by contradiction or by the contrapositive.
Also, note the hypotheses. There are two of them: the sequence $(a_n)$ is bounded,
and any convergent subsequence converges to $a$.
When we see that the sequence is bounded, the first thing that comes to mind
is Bolzano--Weierstrass: any bounded sequence has a convergent subsequence.
But if we compare this with the second hypothesis, it's not so obviously useful: how will it help to apply Bolzano--Weierstrass to try and get $a$ as the limit, when already by hypothesis every convergent subsequence already converges to $a$?
This suggests that it might be worth trying to find an argument where we get
to apply Bolzano--Weierstrass in such a way that we get a convergent subsequence with a limit different from a; and so obtain a contradiction.
In other words, combining the given hypotheses with our basic tool (Bolzano--Weierstrass), it seems that going for a proof by contrapositive/contradiction
might be a fruitful approach. (Of course, we don't know for sure until we're
done $\ldots$; but I'm trying to describe a thought process which would suggest
the approach via the contrapositive.)
Contrapositive also suggests itself if we just write down what we are trying to do:
We want to show that $(a_n) \to a$.
If it does, great! We're done.
If not, then by definition of convergence to $a$, there is an $\epsilon_0 > 0$ and a subsequence
$(a_{n_k})$ of $(a_n)$ such that $|(a_{n_k}) - a| \geq \epsilon_0 $ for all $k$. (Deriving this step is the content of the paragraph
that begins "Suppose ... " and finishes "Stated in plainer language ... ",
together with the first sentence of the next paragraph.)
When you're trying to construct a proof, and you don't see what to do yet, it's always a good idea to write down the definition of what you're trying to prove, and its negation.
Then you can look at them and see which one meshes better with your given
hypotheses.
Here, as I said, there's no obvious mechanism to directly derive that
$(a_n) \to a$. But if we assume the negation, by its very definition it
gives us a subsequence, and we can apply Bolzano--Weierstrass to this
subsequence. (You might ask why we don't apply Bolzano--Weierstrass
directly to our original sequence. But as I already wrote above, this won't
give anything. But our subsequence $(a_{n_k})$ can't converge to $a$ --- it always stays a positive distance $\epsilon_0$ away from $a$. So now we can
get some mileage from Bolzano--Weierstrass.)
As to why we get a sub-sub-sequence, that just comes up naturally when
we apply Bolzano--Weierstrass (which produces a subsequence) to the
negation of the statement that $(a_n) \to a$ (which already prodced a
subsequence).
You shouldn't focus too much on the iterated sub's! Instead, think
in terms of procedures: first we applied the negation of the statement
to be proved (and that gave us a subsequence), then we applied Bolzano--Weierstrass (and that gave us a subsubsequence).
You may want to look at some of Timothy Gower's posts on proofs in real analysis. (My discussion here is to quite a large extent inspired by his
view-point. The link is to a fairly recent post, but he has some earlier posts
too; see here for some. In particular, I like this one. With some googling you should be able to
find more, I think; he is very interested in theorem-proving as a process,
and has written a lot about it in a way that might help you.)
This statement:
Since |$a_n-S$| is bounded, we know that there exists a number, $x\in M$ such that $S - \epsilon \leq x < S$ for all $\epsilon$ >0.*
is stated incorrectly. In fact, there can be no such $x$ since it would satisfy $S \le x < S$.
Prior to that you essentially had the right answer. For each $n$ you can find $a_n \in M$ satisfying $S - \frac 1n < a_n \le S$, from which it follows $|a_n - S| < \frac 1n$ for all $n$.
Given $\epsilon > 0$ choose $N$ satisfying $\frac 1N < \epsilon$. Then $$n \ge N \implies |a_n - S| < \frac 1n \le \frac 1N < \epsilon.$$ This is the definition of $a_n \to S$.
Best Answer
The intuition behind the proof is this. The strict inequality first of all is very important. The result need not be true if the condition is $\sup A \le \sup B$. If $\sup A \lt \sup B$ what you need to see is there are real numbers (uncountably many in fact) between $\sup A$ and $\sup B$. Why because $r = (\sup B - \sup A ) \gt 0$ and forms an interval on the line. Think what happens if none of these real numbers are in $B$. Then none of the elements that are greater than $\sup A$ are in $B$. That is there are no elements in $B$ which are greater than $\sup A$. Then $\sup A$ is an upper bound for $B$. Can this be possible. NO!! Why? Since $\sup B$ is the least upper bound of $B$ and $\sup A$ is a value less than it. This is the story in layman's terms.
I should add what the above contradiction proves. It says there are numbers in $B$ which are greater than $\sup A$ and hence are an upper bound for $A$.
Here is a Lemma that I have always found useful in understanding the intuition behind suprema. $u = \sup A \iff x \in A \implies x \le u $ and $b \lt u \implies \exists \ x' \in A$ such that $b \lt x'$. This particular Lemma makes a lot of proofs simple.
A direct application for your question states that $\exists b \in B$ such that $ b \gt \sup A$. End of Story!