Error in the proof for Let $A$ and $B$ be sets, then $A^{\complement}\cup B^{\complement}\subseteq (A \cap B)^{\complement}$ I have a proof that I know has a fundamental error in it, but I have no idea where the error is. Here is the proof:


  Proof: Suppose that $x \in A^{\complement} \cup B^{\complement}$. This implies that $x \in A^{\complement}$ or $x \in B^{\complement}$. This disjunction will be true if $x \in A^{\complement}$. In this case, we have that $x \notin A$. But if $x \notin A$ then $x \notin A \cap B$, since $A \cap B$ is the set of all elements that are in both $A$ and $B$. And if $x \notin A \cap B$, then $x \in (A \cap B)^{\complement}$. Therefore, $A^{\complement}\cup B^{\complement}\subseteq (A \cap B)^{\complement}$ as desired.
There is no fundamental error in it. 

You proved legally that $x\in A^{\complement}\implies x\in(A\cap B)^{\complement}$ so that $A^{\complement}\subseteq(A\cap B)^{\complement}$. 

What lacks somehow is a statement as: 

"...likewise we prove that $B^{\complement}\subseteq(A\cap B)^{\complement}$ so that $A^{\complement}\cup B^{\complement}\subseteq(A\cap B)^{\complement}$".
You have not taken into account the fact that you can have both $x\in  A^{\complement}\cup B^{\complement}$ AND $x\in A$... 

Since $ A^{\complement} \cup B^{\complement}$ is the addition of $A^{\complement}$ and $B^{\complement}$, so you can add back elements that were excluded from $A$ but are in $B^{\complement}$.