My proof of Turán's Theorem

The beautiful Turán's Theorem says that the unique graph that maximizes the number of edges in a $n$-vertex graph that does not contain $(r+1)$-clique is a complete $r$-partite graph with $r$ parts of equal or almost-equal size, i.e. no two parts differ in size by more than $1$.

The following proof reflects the order of my thought.

Proof:

Let $G$ be a graph with maximum number of edges while avoiding $(r+1)$-clique.

Step 1
Clearly when restricted to $r$-partite graph, $G$ has all parts of equal or almost-equal size -- if not, move a vertex from a smaller part to a larger one and the edge count increases. It remains to show that $G$ has to be $r$-partite.

Step 2
Inspired by $r$-partite graph, if we can prove

Lemma
$G$ doesn't have vertices $u, v, w$ such that only $u$ and $v$ among them are adjacent

Then the theorem is proved. This is because we could start with a $r$-clique $C\subset G$ as the $r$ parts $P_1,\ldots,P_r$, and place each other node $v\in V(G)-C$ in part $P_i$ associated with $c_i\in C$ such that $v$ and $c_i$ are not adjacent. By the Lemma, each part is now an independence set and therefore $G$ is $r$-partite.

Step 3
Also inspired by complete $r$-partite graph, every two non-adjacent nodes $u$ and $v$ in $G$ must have the same degree. If not say $d(u)\gt d(v)$. By the operation of deleting all edges incident to $v$ and make it connect to the same set of vertices as $u$, we obtain a graph with more edges while still free of $(r+1)$-clique.

Step 4
To prove the Lemma, say among vertices $u, v, w$ only $u$ and $v$ are adjacent. We could still perform the operation even when $d(w)=d(u)=d(v)$: remove all edges incident to $u$ and make it link to the same vertices as $w$. Now the new graph $G'$ is still optimal, yet $d(u)=d(w)=d(v)+1$, contradiction.

Q.E.D.