To prove 4-SAT CNF is NP-complete
In order to prove that 4-SAT is NP-complete, you need to prove that it is in NP and that it is NP-hard.

Prove 4-SAT $\in$ NP

Given an instance of 4-SAT and an answer that evaluates to TRUE, it's pretty quick to verify.

4-SAT is NP-hard

Like the comments suggested, reduce 3-SAT, which is a known NP-complete problem, to 4-SAT:

Assume that you are given an instance of 3-SAT. The goal is to convert it in polynomial time into an instance of 4-SAT in such a way that the answer is preserved.

Take each clause $(x \lor y \lor z)$ and turn it into $(x \lor y \lor z \lor a) \land (x \lor y \lor z \lor \neg a)$, where $a$ is arbitrarily set.


If any $(x \lor y \lor z)$ clause is satisfied, then $(x \lor y \lor z \lor a) \land (x \lor y \lor z \lor \neg a)$ will also be satisfied. If the instance of 3-SAT is satisfiable, then the new instance of 4-SAT is satisfiable.
If $(x \lor y \lor z \lor a) \land (x \lor y \lor z \lor \neg a)$ is satisfied, then $(x \lor y \lor z)$ must also be true because $a$ is opposite of $\neg a$. If the new instance of 4-SAT is satisfiable, then the instance of 3-SAT is satisfiable.