So I was thinking about going down a couple different roads but none seemed to work.
First of all I was thinking about using arithmetica of continuous function and looking for a limit on some $x_0 \in [0,1]$ but I'm not quite sure.
I know that since h is arbitrary then basically I can show $\lim h = 0$ but I am kind of lost as to how I can show this formally.
Another thing I was thinking was defining a function such as $g(x) = f(x) - (x+h)$ but that also didn't seem to work.
Anyway I would love some help!
EDIT:
I have sketched some proof and I would like to see if it is thorough enough:
Let us look at the helper function $g(x) = f(x) - x$, $g:[0,1] \to \mathbf R$.
From the datum $f(x)$ is continuous in $[0,1]$. Also, denote x is an elementarien function, thus continuous in $[0,1]$.
By arthmetica of continuous functions we have $g(x)$ is continuous in $[0,1]$.
From Weir Strauss theorem every continuous function in a closed segment has a minimum and a maximum. Thus let $x_0 \in [0,1]$ denote the minimum of $g(x)$ in $[0,1]$.
Denote, $g(x_0) \le g(x), \forall x \in [0,1]$.
From the datum we have $f(x) \gt x, \forall x \in [0,1]$ and specifically we have $f(x_0) - x_0 \gt 0$, thus implementing from the density of irrationals we have $h>0$ such that:
$g(x) \gt g(x_0) \gt h \gt 0$.
Thus meaning we have $g(x) > h, \forall x \in [0,1] \implies f(x) \gt x + h, \forall x \in [0,1]$.
As needed.
Let $g=f-x$, so $g>0$ and is continuous on this compact set $[0,1]$, hence can attain its maxima and minima. The minima can't be zero, otherwise contradicts with $g>0$, hence the minima is some positive number, define it as: $\min g=a>0$. Now take $h=a/2$ and it is done.