I've read the Intermediate Value Theorem's proof and I understand everything except one line which i've been speculating how it is derived. I know this may be a silly question but it does confuse me. So I will write down the proof from the book and say which line I don't understand:
IVT:
Let $f : [a, b] → \mathbb R$ be continuous and suppose that $u$ lies between $f(a)$ and $f(b)$. Then there is a point $c$ between $a$ and $b$ where $f(c) = u$.
PROOF: Assume that $f(a) < u < f(b)$ and let $A$ be the set $\{x \in [a, b] : f(x) \leq u\}$.
This set is non-empty since it contains $a$ and is bounded above by $b$. Let $s$ be its least upper bound. The aim is to show that $f(s) = u$.
Suppose that we have $f(s) < u$.
Note that that $s \neq b$ because we know $f(b) > u$.
If we put $E = u − f(s)$ then for some $δ > 0$ $|f(x) − f(s)| < E$ as long as $|x − s| < δ$.
So in particular if $x = s + δ/2$ then $f(x) < f(s) + E = u$. (HERE: I do not understand completely why $x=s+/2$. Is it because it satisfies the condition of $>0$ for when we substitute it in the equation $|x-s|< $??)
This means that $s + δ/2 \in A$ contradicting the fact that $s$ is an upper bound for $A$.
I know the proof continues but I understand the rest.
Yes. We want to show that $s+\delta/2\in A$, or equivalently $f(x)\le u$ when $x=s+\delta/2$. So we check that $x=s+\delta/2$ satisfies $|x-s|<\delta$, and thus $|f(x)-f(s)|<E$, which implies $f(x)\le E+f(s)=u$.