I am assuming that the intermediate value theorem you have studied, must be somewhat like this :
Given a continuous function $f : [a,b] \to [c,d]$, for all $e \in [c,d]$ there exists $g \in [a,b]$ such that $f(g) = e$.
Now, in our case, we have a function $f$ from $[0,1] \to [0,1]$. What we do is to create a new function out of this one, and then apply the IVT on that. Note this procedure carefully, you wil see it many times.
Let $g(x) = f(x) - x$. Then, what is the range of $g$? Note that the smallest and largest value of both $x$ and $f(x)$ are $0$ and $1$ respectively.Hence, the largest value of the difference is $1$, and the smallest value is $-1$.
Hence, $g : [0,1] \to [-1,1]$. By IVT, since $0 \in [-1,1]$, we know that there is some $h \in [0,1]$ such that $g(h) = 0$. But what is $g(h)$? It is $f(h)-h$, by definition! Hence, we get $f(h) -h= 0 \implies f(h)=h$, for some $h \in [0,1]$.
This proof did not go through any $\epsilon-\delta$, howver the proof the IVT itself goes through that. If you want an explanation as to how the IVT is proved (how these $\epsilon$s and $\delta$s work with each other) then I can help you.