Show that if $f:[a,b]\rightarrow \mathbb{R}$ is continuous and $[a,b]\subseteq f([a,b])$ then there exists a fixed point $f(x)=x$ I have tried to prove the statement but there is a part where I am not sure if I can argue like that, I will mark it by setting it in bold
We know that $[a,b]\subseteq f([a,b])$ and that $f$  is continuous
$$\Longrightarrow \exists_{x,y\in [a,b]}f(x)=a\wedge f(y)=b$$
Without loss of generality assume $x<y$
$$\Longrightarrow[x,y]\subseteq[a,b]$$
The function $f_{|[x,y]}$ is continuous and for all $\gamma\in [f(x),f(y)]$ there is a $z\in[x,y]$ such that $f(z)= \gamma$ 

Here is the part where I am not quite sure. Also if you have an idea how to prove it differently please share it with me:
Assume there exists finitely many points $z\in [x,y]$ such that $f(z)\neq z$.
Then we are already done because that means there must exist infinitely many Points for which the opposite is true, since the interval consists of infinitely many Points.
Therefore assume there exists infinitely many Points such that $z\in [x,y] \wedge f(z)\neq z$
Then there must also exist infinitely many points for which $f(z)>z$ and $f(z')<z'$
Without loss of generality assume there are only finitely many Points for which $z<f(z)$. Then there exists a maximal element of this set and one can split the interval $[x,y]$ into $[x,z]$ and $(z,y]$. If there exists a $w\in (z,y]$ such that $f(w) = w$ we are already done. Therefore assume $\forall_{w\in (z,y]} f(w)>w$. This would invoke a contradiction that $f$ is continuous in $z$.
Therefore we have showed there exists infinitely many $z,z'\in [x,y]$ such that $f(z)>z$ and $f(z')<z'$ 
One can pick now $z,z'$ such that $f(z)>z$ and $f(z')<z'$
Without loss of generality $z<z'$
Then $[z,z']$ is a closed interval and due to the Intermediate-value-theorem
$$\forall_{\gamma\in[f(z),f(z')]}\exists_{x\in[z,z']}f(x)=\gamma$$
If we repeat this process by Always choosing an element $l$ for the left end of the interval for which we have $f(l)>l$ and for the Right end an element for which we have $f(r)<r$. Then we get a nested interval which must converge to a fixed Point.

Is my thought process consistent? One Thing that this idea lacks is probably that there must always exist Points $l$ and $r$ like described above otherwise the proof does not work. Is it possible to prove this somehow?
 A: Assume that there is no fixed point, $f(x)\ne x$ for all $x\in\mathbb{R}$. Then, by continuity, either $f(x)>x$ for all $x\in\mathbb{R}$, or $f(x)<x$ for all $x\in\mathbb{R}$. Assume the former. Then, for each $x\in[a,b]$, one has $f(x)>x\ge a$, that is, $f(x)>a$. Hence $a\notin f([a,b])$, and we are done. The second case can be considered in a similar way. 
A: There is no reason to split your argument into counting finitely/infinitely many things. The moment you have a single fixed point, you're done. Or if you don't have any fixed points, then you will have at least one(infinite or not doesn't matter) $z$ and one $z'$ (guaranteed by the assumption that $[a,b]$ is contained in the image) such that

$f(z)>z$
   and $f(z′)<z′$

Once you have this, you can just apply intermediate value theorem once as in the standard proof (see below). 
PS I cannot understand what you wrote starting from the line

Without loss of generality assume there are only finitely many Points for which $z<f(z)$.


The standard proof. Without loss $[a,b]= [0,1]$. $f$ is continuous, so by extreme value theorem, it attains its min $m\le 0$ and max $M\ge 1$ at the points $x_0$ and $x_1$ respectively. Define for $t\in[0,1]$, $x_t = tx_0 + (1-t)x_1$ and consider $F(t) = f(x_t)-x_t$. We have $F(0) = m - x_0 ≤  -x_0\le 0$, and $F(1) = M - x_1 ≥ 1-x_1 \ge 0.$ By intermediate value theorem, there is a point $s$ in $[0,1]$ such that $F(s) = 0$. This corresponds to a point $x_s \in [0,1]$ such that $f(x_s) = x_s$, as needed.
