Prove that finding a cliques is NP-Complete? I presented my proof bellow, so is it complete (formal) proof.
Proof :
1- We can verify the solution in Polynomial time.
2- 3-SAT (NP-Complete) can be polynomialy reduced to clique! (as the following):


*

*We assume a formula 3-sat (F) as an example :


F= (X1 + X2 + X3).(^X1 + ^X2 + X3).(X4 + ^X3 + X2)
note: ^X is the negation of X
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3-SAT is satisfiable $<=>$ $G(V,E)$ has a clique of size k>0.
1 Clique =>3-SAT : if we have a clique then it is SAT "by construction" literal represented as Node. such that no edge between $X_i$,$X_j$  belongs $C_k$, and no edge  between ^$X_i$ and $X_i$   in both $C_k, C_j$.
let  G has a clique of size k we can set each literal represented by a node v belongs the clique so 3-SAT is there.
since it is enough to have only 1 literal  = true in each clause.
[2] 3-SAT => Clique:
Let x be 3-SAT found in CNF, and it is satisfiable. => by the graph construction- we can connect all nodes that have true for each clause, and hence we have a clique #
Is my proof correct ? 

Your proof is correct and your steps are clear.
