Prove that there cannot exist a set that contains everything.
Ill put my proof in the answer so please check it there. Also if there is a more creative way to do this(using the basic axioms) if it's not too much a bother can you please share that. Thank you.
On
Proof.
We begin by using the axiom of specification with the formula $(x\notin x)$ on $A$. If it true that
$$\forall y(y\in B \leftrightarrow(y\in A \wedge (y \notin y)),$$
then so is,
$$B\notin A$$
Assume the $B\in A$. For every element $x$ in $A$, either $x\in B$ or $x\notin B$ and the same is true for $B$, that is either $B\in B$ or $B\notin B$.
Case 1: If $B\in B$, then this is a contradiction because $B$ is the set of elements that do not contain themselves.
Case 2: If $B\notin B$, then $B$ is an element of $A$ such that it is does not contain itself therefore it must also be member of $B$, which is also a contradiction.
This leads us to the fact that the original assumption, that is $B\in A$, is false. Therefore it must be $B\notin A$.
Now this means there exists an arbitrary set$A$ such that it's subset $B$ cannot be contained in $A$.
My favorite is the following:
Let $A$ be a set containing all sets. Then it is easy to see that $\mathcal{P}(A)\subseteq A$. Then $|\mathcal{P}(A)|\le A$, a contradiction of Cantor's Theorem.