Assuming: $\forall x \in [0,1]:f(x) > x$ Prove: $\forall x \in [0,1]:f(x) > x + \varepsilon $

Let $f$ a continous function defined in the interval $[0,1]$.
Assuming: $\forall x \in [0,1]:f(x) > x$
Prove: $\forall x \in [0,1]:f(x) > x + \varepsilon $

I tried to use Heine–Cantor theorem and did some Algebra tricks but it didn't bring me to a safe shore :)

What do you suggest?
Thanks!

Let $h:x\mapsto f(x)-x$ then $h$ is continuous on the compact $I=[0,1]$ hence $h$ is bounded (here we need to say bounded below) and reaches its minimum at $x_0\in I$ hence we have $$h(x)\geq h(x_0)>0,\;\forall x\in I$$ Now take $\epsilon=\frac{h(x_0)}{2}$ and you have the result.

Let $g(x) = f(x) - x$, then for each $x\in [0,1]$, $g(x) > 0$, so for $\epsilon_x := g(x)/2$, there is a $\delta_x > 0$ such that $$ y \in (x-\delta_x,x+\delta_x) \Rightarrow g(y) > \epsilon_x $$ Now $\{(x-\delta_x, x+\delta_x)\}$ is an open cover for $[0,1]$. Take a finite subcover, and let $\epsilon$ be the minimum of all the corresponding $\epsilon_x$'s.