Lemma 1.2.7. Among $n$-vertex $r$-partite graphs, $T_{n,r}$ is the unique graph with the maximum number of edges.
Theorem. The Turan graph $T_{n,r}$ maximizes the number of edges among all $n$-vertex $K_{r+1}$-free graphs. It is also the unique maximizer.
Proof. As before, let $G$ be an $n$-vertex, $K_{r+1}$-free graph with the maximum possible number of edges.
We claim that if $x$ and $y$ are non-adjacent vertices, then $\deg x=\deg y$. Indeed, suppose $\deg x>\deg y$. We can modify $G$ by removing $y$ and adding in a clone of $x$ (a new vertex $x'$ with the same neightborhood as $x$ but not adjacent to $x$), as illustrated below.
The resulting graph would still be $K_{r+1}$-free (since a clique cannot contain both $x$ and its clone) and has strictly more edges than $G$, thereby contradicting the assumption that $G$ has the maximum possible number of edges.
Suppose $x$ is non-adjacent to both $y$ and $z$ in $G$. We claim that $y$ and $z$ must be non-adjacent. We just saw that $\deg x=\deg y=\deg z$. If $yz$ is an edge, then by deleting $y$ and $z$ from $G$ and adding two clones of $x$, we obtain a $K_{r+1}$-free graph with one more edge than $G$. This would contradict the maximality of $G$.
Therefore, non-adjacency is an equivalence relation among vertcies of $G$. So the complememnt of $G$ is a union of cliques. Hence $G$ is a complete multipartite graph, which has at most $r$ parts since $G$ is $K_{r+1}$-free. Among all complete $r$-partite graphs, the Turan graph $T_{n,r}$ is the unique graph that maximizes the number of edges, by Lemma 1.2.7. Therefore, $G$ is isomorphic to $T_{n,r}$.
I understand the general idea of the proof but I'd like to clarify some moments.
Question 1. Assume that $V(G)=\{v_1,\dots,v_n\}$. The relation $v_i\sim v_j \Leftrightarrow v_iv_j\notin E(G)$ is an equivalence realtion on $V(G)$. Since any equivalence relation on set forms its partition, then $V(G)=V_1\sqcup \dots\sqcup V_N$, where $V_i$ are equivalence classes. More precisely, each $V_i$ is an independent set and for any $a\in V_i$, $b\in V_j$ with $i\neq j$ there is an edge connecting $a$ and $b$. It means that $G$ is complete $N$-partite graph. Is my reasoning correct here? I do not see the point why the author says that the complement of graph is a bunch of cliques.
Question 2. We know that initially $G$ is $n$-vertex, $K_{r+1}$-free graph with the maximum number of edges and we were able to show $G$ is complete, $N$-partite graph with $N\leq r$. I am confused how the author used the Lemma to show that $G=T_{n,r}$. How to show that $N=r$ and only in this case it has the maximum number of edges?
I'd be really thankful for help! Thank you so much!
Question 1) Yes. You are correct. It's maybe a bit easier visually to think of cliques as being equivalence classes (and is a common example to think about), so that may be why the author chose to phrase it in terms of complement.
Question 2) The main theorem proof shows that $G$ is $r$-partite. As pointed out by jacob in the comments - this means it has at most $r$ maximal independent sets. As you've noticed, it could in theory have fewer from the proof thus far.
What the Lemma is really saying, with this understanding of what $r$-partite means is: among all $n$-vertex complete multipartite graphs with $r$ or fewer partite sets, $T_{n,r}$ is the unique edge maximiser. We know $G$ is $r$-partite with $n$ vertices, and maximises the edge count, so it is $T_{n,r}$ by the Lemma. I.e., the Lemma tells you $N=r$.