Confusion regarding the proof of Theorem I.32 of Apostol's Calculus. On page 27 of his book "Calculus Volume One", Apostol writes the following theorem and its proof:

Theorem I.32. Let $h$ be a given positive number and let $S$ be a set of real numbers.
(a) If $S$ has a supremum, then for some $x$ in $S$ we have $$x > \sup S - h.$$
(b) If $S$ has an infimum, then for some $x$ in $S$ we have $$x < \inf S + h.$$
$\textit{Proof of } (a).$ If we had $x \le \sup S - h$ for all $x$ in $S$, then $\sup S - h$ would be an upper bound for $S$ smaller than its least upper bound. Therefore we must have $x > \sup S - h$ for at least one $x$ in $S$. This proves (a). The proof of (b) is similar.

My problem is that I don't intuitively understand the proof of a, nor am I sure how to write the proof of b. I am asking in my endeavors to self-study Calculus. Any help you could provide would be most helpful.
 A: The intuition behind a) is the following: if a number $A$ is the least upper bound of the set $S$, i.e. $A=\sup S$, then whatever small $h>0$ you pick, and search within the interval $(A-h,A)$, you will for sure find an element of $S$ there. In other words, there must be elements of the set $S$ within every small distance below $A$. It is because one cannot move $A$ any further to the left to make it smaller - something (the set $S$) must prevent doing it. If you can move $A$ to the left so that it remains to be an upper bound then this upper bound is not the least.
To prove b) you just need to make a symmetric counterpart of this for the largest lower bound. 
A: We’ll use the fact that the supremum or least upper bound of a subset $S \subset \mathbb{R}$ is an element of $\mathbb{R}$ so that every element of $S$ is less then or equal to this distinguished element of $\mathbb{R}$ and that this is the smallest such element. Now we will argue using a proof by contradiction.
Let us suppose that we are given a positive number $h$. We want to show that there is some $x \in S$ such that $ x > \sup S -h$. Suppose to the contrary there is no such $x$. That means that for all $x \in S$, we have that $x \leq \sup S - h$. So $\sup S - h$ is an upper bound for the set $S$. Notice that $\sup S - h < \sup S$ since $h$ is positive. This contradicts the fact that $\sup S$ is a least upper bound. So our assumption was wrong and so there is some $x \in S$ so that $ x > \sup S - h$.
B) Is exactly the same proof with the roles of upper and lower bounds switched. Assume the opposite and show that you end up with a lower bound greater than the infimum. Hope this helps!
A: A proof for b): Suppose you had $\inf{S}+h<x,\forall x\in S$. Then $\inf{S}+h$ would be a lower bound for the set $S$, but greater than $\inf{S}$, which is by definition (and we know that it exists, why?) the greatest lower bound. That is, we would have a lower bound greater than the greatest lower bound, which is a contradiction! So there must exist a $x\in S$ such that $\inf{S}+h>x$. The intuition behind this theorem is that if you "nudge" the infimum up, or the supremum down a little bit, you already "inside" your interval.
