The Theorem is
Let $r \geq 1$ be an integer and let $\epsilon > 0$.
Then there is an integer $n_0 = n_0(r_1\epsilon)$ such that if $|G| = n \geq n_0$ and $$\delta (G) \geq \left( 1-\frac{1}{r} +\epsilon \right)n,$$ then $ K_{r+1}(t) \subset G$, where $$t \geq \frac{\epsilon\text{log}(n)}{2^{r-1}(r-1)!}.$$
I have taken this statement of the theorem from Modern Graph Theory by Bollobas, from which I am also trying to understand the proof. I have referred to other lecture notes that include the proof here:
However, I am stuck right at the beginning. (Perhaps I will get stuck again!)
All sources use an inductive proof on $r$. I will go by the proof in Bollobas.
The case for $r=1$ is related to the Zarankiewicz problem of finding the maximum size of an $n \times n$ bipartite graph not containing $K_{t,t}$, and has been proved already. There is also a weaker but sufficient-for-the-purposes demonstration.
Given the $r=1$ case, take $r\geq 2$ and let $G$ be a graph with $n$ vertices and minimal degree at least $(1-\frac{1}{r} +\epsilon)n$. Note that $0<\epsilon<1/r$.
Since $\delta(G) > (1-\frac{1}{r}+\frac{1}{r(r-1)})n$, by the induction hypothesis $G$ contains a $K_r(T)=K$, say, with $|T|=[d_r \log(n)]$.
I do not understand the first part of that statement. Why is $\delta(G) > (1-\frac{1}{r}+\frac{1}{r(r-1)})n$? I do not see anything that has been said which implies this to be the case? Since we only have $\epsilon >0$, then strictly speaking we could just have $0$ here!
This is a typo. The corrected argument is that since $$ \delta(G) \ge \left(1 - \frac1r\right)n = \left(1 - \color{red}{\frac1{r-1}} + \frac1{r(r-1)}\right)n $$ we have $\delta(G) \ge \left(1 - \frac1{r-1} + \epsilon'\right)n$ for $\epsilon'= \frac1{r(r-1)}$. So we can, by induction, apply the $r-1$ case of this theorem to conclude that $G$ contains a $K_{r}(T)$, where $$ T \ge \frac{\epsilon'\log n}{2^{r-2}(r-2)!} = \frac{\frac1{r(r-1)} \log n}{2^{r-2} (r-2)!} = \frac{\log n}{2^{r-2} r!}. $$ This expression is rewritten as $d_r \log n$, where $d_r = \frac{2^{2-r}}{r!}$, in the next step.