real analysis - Starting Bisection Proof of Extreme Value Theorem

I am having difficulties beginning a proof for the following statement:

Use a proof strategy of bisection to prove that every function $f:[a,b] \to \mathbb{R}$ that is not bounded above is discontinuous at some point $c \in [a,b]$ (and discontinuous from the right or left if $c=a$ or $b$, respectively.

Although the strategy is given, I am having trouble getting started. Furthermore, how would I use this bisection argument to prove that if $f:[a,b] \to \mathbb{R}$ is continuous and $\sup{\{f(x):a \leq x \leq b\}}=M$, then $f(c)=M$ for some $c \in [a,b]$. I believe that this is a reformulation of the Extreme Value Theorem.

Many thanks in advance. I am using the textbook Introduction to Analysis by Arthur Mattuck.

Answer

If $f:[a,b] \to \mathbb{R}$
is not bounded above,
there is a point $x_1 \in [a, b]$
such that
$f(x_1) > 1$.

Divide $[a, b]$
into two parts
$[a, x_1]$ and $[x_1, b]$.

In one of these parts,
there is an $x_2$
such that
$f(x_2) > 2$
(by the unboundedness of $f$).
The width of this interval
is $\le \frac{b-a}{2}$.

Again,
divide that part

at $x_2$ into two parts.
The width of each of these parts
is $\le \frac{b-a}{4}$,
and there must be an
$x_3$ in one of these parts
such that
$f(x_3) > 3$
(or $f(x_3) >$ some large value if you want).

Repeating this,

after the $n$-th division,
there is an interval
or width $\le \frac{b-a}{2^{n-1}}$
with a point $x_n$
such that
$f(x_n) > n$.
This sequence of points
$(x_n)$
converges to a limit
and $f(x_n)$

is unbounded,
so $f$ can not be continuous
at the limit.