Prove $\mathit f(x)=x$ for some $x\in\mathbb R$. Let $f:\mathbb R\to\mathbb R$ be monotonically increasing (perhaps discontinuous). Suppose $f(0)\gt 0$ and $f(100)\lt 100$. 
Prove $\mathit f(x)=x$ for some $x\in\mathbb R$.

Here are my thoughts. I tried to use the way of contradiction. 
Suppose there is no fixed point. Then $f$ can not be a contraction. Then $\forall\alpha\in(0,1)$, $\exists x_1$, $x_2$ such that $d\bigl(f(x_1),f(x_2)\bigl)\gt \alpha d(x_1,x_2)$.
However, I can only extract things below from the given conditions, which does not contribute to my "way of contradiction" method at all, in my opinion.
Since $f$ is monotonically increasing, $f(100) \gt f(0)$. 
Since $f(0)\gt 0$ and $f(100)\lt 100$, $d\bigl(f(0),f(100)\bigl)=|f(0)-f(100)|\lt 100\lt |0-100|=d(0,100)$.

I think I should definitely somehow use this but I currently have no idea how. 
Could anyone help me figure this out?
 A: Let $B = \{y| 0\le y \le 100; f(y) \le y\}$.  $100 \in B$ and $B$ is bounded below by $0$.  So $\inf B$ exists and $\inf B \ge 0$.
If $0 < x< f(0)$ then $f(0) < f(x)$  because $f$ is mono increasing so $x < f(0) < f(x)$ so $x\not \in B$ so $f(0)\le \inf B$.
So what is $f(\inf B)$?
If $f(\inf B) > \inf B$ then there is a $y; \inf B < y <f(\inf B)$ so that $y\in B$.  Because $y >\inf B$ and $f$ is monot increasing $f(y) > f(\inf B) > y$.  Thus $y\not \in B$.  This is a contradiction.
If $f(\inf B) < \inf B$ then there is a $\max(0, f(\inf B))< y <\inf(B)$. $y\not \in B$ so $f(y)> y$.  But now we have $y< \inf(b)$ but $f(y) > y >f(\inf B)$ violating $f$ being increasing.
So $f(\inf B) = \inf B$
======== old version ======
I suppose I'd let $A = \{x\ge 0|$ for all $y;0\le y \le x; f(y)>y\}$.  It's easy to see that $A$ is an interval and that $[0,f(0)] \subset A \subsetneq [0,100]$ and so $A$ is bounded above by $100$ and so $\sup A$ exists.
Pf: If $A$ weren't an interval then there would be an $0\le x < y$ so that $x\not \in A$ and $y \in A$ so there would be an $a \in [0,x]$ so that $f(a)\le f(a)$ but then, as $a\in [0,y]$ we have $y \not \in A$.  And $f(0) > 0$ so $0\in A$.  And $f(0) > 0$ so as $f$ is mono increasing $f(f(0)) >f(0)$ so $f(0)\in A$ and so $[0,f(0)]\subset A$ and as $f(100) < 100$, $100 \not \in A$ so $A\subsetneq [0,100]$.
Okay so what is $f(\sup A)$?
$f(\sup A) < \sup A$ is impossible.  If $f(\sup A) < \sup A$ then let $\epsilon = \sup A - f(\sup A)$.  Then there is a $y \in A$ so that $\sup A - \epsilon <  y \le \sup A$ and $f(y) > y >\sup A - \epsilon = f(\sup A)$ which violates $f$ being mono increasing.
And $f(\sup A) > \sup A$ is impossible. If so, That would mean $\sup A \in A$. Let $z$ be so that $\sup A < z < f(\sup A)$.  Then $z\ne A$ so there an $a\in [0, z]$ so that $f(a) \le a$ but $a \not \in [0,\sup A]\subset A$ so $a > \sup A$. But $f(a) \le a < f(\sup A)$ which violates that $f$ is increasing.
So $f(\sup A) = \sup A$.
A: Consider the function $g(x) = f(x) - x$. Finding a zero of $g$ is equivalent to finding a fixed point of $f$ (but it is easier to visualize).
Now, as mentioned in the comments the claim is not true as stated (taking $g(x) = -1$). Instead, I'm going to assume you meant $f(0) > 0$ and $f(100) < 100$. Under the reverse assumptions that $f(0) < 0$ and $f(100) > 100$, the claim is also not true. For instance, take
$$
g(x) = 
\begin{cases}
  x - \epsilon & \mathrm{if} \ x \leq 0 \\
  x + \epsilon & \mathrm{if} \ x > 0
\end{cases}
$$
for some $\epsilon > 0$.
As for hints, look at the different intervals $x < 0$, $x > 100$ and $0 < x < 100$. Try either to prove that no fixed point can occur in an interval, or construct an example with none. You can patch together these results into a full proof by contradiction.
I've put a worked out version in spoilers, but I can't get the syntax quite right, thus the letter b splitting the paragraphs.

 For $x < 0$, you can see that $f(x) < 0$, so $g(x) < x$.
 Therefore the zero of $g$ cannot come for $x < 0$.
 However, note that $g(100) < 0$ by assumption, so the question is whether the increase in $g$ is enough to bring it to zero.
 Not necessarily; we could have that restricted to $x > 100$, $f(x) = f(100) + (x - 100)$ or in other words $g(x) = g(100) < 0$.

b

 That is, it is possible to construct $f$ with no fixed point outside the interval $[0, 100]$.
 However, there it is necessary for $g$ to be generally decreasing, since we have $g(0) > 0$ and $g(100) < 0$.
 If it decreases in a continuous way, there will be a zero and thus a fixed point of $f$.
 Thus it must have a jump discontinuity, particularly a jump downward (a jump upward would do nothing to avoid the problem as the same logic would apply to the smaller interval ahead of it).

b

 Such a jump discontinuity cannot occur.
 Why? The assumption that $f$ is  increasing implies that the "slope" of $g$ cannot be less than $-1$.
 More precisely, for any $x, x'$ we cannot have $[g(x') - g(x)]/[x' - x] < -1$.
 For points surrounding the jump continuity, this quantity can be arbitrarily large negative values (intuitively, the slope at a discontinuity is infinite).
 Therefore we have a contradiction: any constructed function $f$ without a fixed point does not satisfy the property.

