Proof continuous function $g$ defined on $[0,1]$ has a fixedpoint $x \in [0,1]$ Claim: For the continuous function $g:[0,1]\rightarrow[0,1] \exists x \in [0,1]: g(x_1) = x_1$
Proof:
Case 1: If g(0) = 0 and/or g(1) = 1 then g has at least one fix point on one of the ends of the intervall. This means $\exists x_1\in[0,1]: g(x_1)= x_1$
Case 2: Suppose $g(0) \ne 0 $ and $g(1) \ne 1 \Rightarrow$ g does not have a fixed point on one of the ends of the intervall. Therefore we have that g(0) > 0 since (0 < g(0) < 1) and g (1) < 1 since (0 < g(1) < 1) now define:
$$h(x) := g(x) -x$$
$\Rightarrow$ h is continuous since it is constructed by adding two continuous functions. Now: 
$$h(0)= g(0)-0>0. \,\,By \,\, g(0)>0$$
$$h(1) = g(1) - 1 < 0. \,\, By \,\, g(1) < 1$$
By the intermediate value thm. $\Rightarrow \exists x_1 [0,1]: h(x_1) = 0$ this implies $\Rightarrow g(x_1)-x_1= 0 \iff g(x_1) = x_1$
Thus $x_1$ is a fixed point of g. By this the claim is proven. q.e.d
Now I'm asking for verification for that proof.

Your proof is correct! 
Geometrically, this says, any continuous function $g$ from $[0,1]$ to itself MUST intersect the diagonal $y=x$, and that intersecting point is the fixed point
