In this case, "Applying induction on $S$" means "Applying the $k-1$ case of the theorem to the subgraph $G[S]$".
Formally, the theorem we're proving here is the following version of Turán's theorem:
Theorem. For all $n,k$, if $G$ is an $n$-vertex graph with no $k$-clique, then there is a $(k-1)$-partite $n$-vertex graph with at least as many edges as $G$.
Going through all the things you want help with:
The construction of $H$. We let $m$ be a vertex in $G$ with the largest degree; $S$ is the set of neighbors of $m$, and $T$ is the complement of $S$ (so $m \in T$). To get $H$ from $G$, we:
- Delete all edges within $T$ (if there were any);
- Add all missing edges between $S$ and $T$ (if there were any).
Why is $|E(H)| \ge |E(G)|$? We prove a stronger statement: for all vertices $v$, $\deg_H(v) \ge \deg_G(v)$. This gives us an inequality between the edge counts, since the sum of degrees gives us twice the number of edges.
We check the degree inequality separately in two cases:
- Suppose $v \in S$. In this case, going from $G$ to $H$, we did not delete any edges out of $v$, but possibly we added some edges (if $v$ was not already adjacent to all of $T$). So we can only increase the degree of $v$.
- Suppose $v \in T$. In this case, the degree of $v$ in $H$ is $|S|$, since $H$ contains all edges between $S$ and $T$. In $G$, the degree of $v$ was at most the degree of $m$ (by our choice of $m$), so it was at mot $|S|$.
How does the induction work? It is an induction on $k$. The base case can be $k=2$: the only graph with no $2$-cliques (edges) is the empty graph, which is a $1$-partite graph.
If $G$ started out as a graph with no $k$-cliques, then the subgraph $G[S]$ must be a subgraph with no $(k-1)$-cliques (any $(k-1)$-clique in $G[S]$ would form a $k$-clique together with $m$). By induction, there is some $(k-2)$-partite graph $K$ on $|S|$ vertices with $|E(K)| \ge |E(G[S])|$.
Now modify $H$ by replacing $G[S]$ (or $H[S]$, which is the same thing) with the $(k-2)$-partite graph $K$. The result (call it $H'$)
- Still has at least as many edges as $G$: the number of edges did not decrease when going from $G$ to $H$, and it did not decrease when replacing $G[S]$ with $K$.
- Is $(k-1)$-partite. Since $K$ is a $(k-2)$-partite graph on $S$, there is a partition $S = S_1 \cup S_2 \cup \dots \cup S_{k-2}$ such that $K$ has no edges inside any $S_i$. Also, $H$ has no edges inside $T$, because we deleted them all. So $H'$ has a vertex partition $S_1 \cup S_2 \cup \dots \cup S_{k-2} \cup T$ with no edges inside any part: this is the definition of a $(k-1)$-partite graph.
This proves the theorem.
For convenience, I'll refer to the second-to-last sentence of the proof you've posted as $\star$.
For each $b\in B$, let $a_b\in A$ be the unique element of $A$ not connected to $b$. Fact $\star$ now says precisely the $a_b\neq a_c$ for any $i\neq j$ and $b\in V_i,c\in V_j$.
Okay, now pick any $b_1\in V_1,\dots,b_r\in V_r$. By $\star$, we have $a_{b_i}\neq a_{b_j}$ for all $i\neq j$, whence $A=\{a_{b_1},\dots,a_{b_r}\}$. Now I claim that $N(b)\cap A=A\setminus\{a_{b_i}\}$ for all $i\in\{1,\dots,r\}$ and $b\in V_i$; this will show the desired result. If it were not the case, then there would be some $i$ and $c\in V_i$ such that $N(c)\cap A=A\setminus\{a_{b_j}\}$ for some $j\neq i$. But then $a_c=a_{b_j}$, contradicting fact $\star$, so we are done.
Once you've shown this, then you're done. Just to spell it out in detail: by the argument above, for each $i\in\{1,\dots,r\}$, there is a vertex $a_i\in A$ such that $a_i\notin N(b)$ for all $b\in V_i$. Let $W_i=V_i\cup\{a_i\}$; then $e(W_i)$ is empty for each $i$ by the definition of $V_i$ and $a_i$. Moreover, we know $G[V_i\cup V_j]$ is complete bipartite for $i\neq j$, and by fact $\star$ we have that $a_i$ is connected to every element of $V_j$ for $i\neq j$. So $G[W_i\cup W_j]$ is complete bipartite, as needed.
Best Answer
Notation: $e(G)$ is the number of edges of a graph $G$. $\overline{G}$ is the complement of $G$. $\omega(G)$ is the clique number of $G$.
Note that $T'(n,q) = \overline{T(n,q)}$, and $\omega(G) = \alpha(\overline{G})$.
$e(\overline{G}) = \binom{n}{2} - e(G) = e(T(n,q)) > e(T(n,q-1))$. By Turan's theorem, $\alpha(G) = \omega(\overline{G}) \geq (q-1)+1$.
Edit.
It implies that
Of course, the dual version implies the original. However,
I think it is quite weaker (as you said) to prove the original Turan's theorem since $e(T'(n,q)) < e(T'(n,q-1))$.
In addition, I think that you miss '$G$ minimizing its edges' in this sentence, "additionally, if $\alpha(G)=q$ then it must be that $G\cong T'(n,q)$."