I'll rewrite my answer.
First, the question is vague. You are saying:
I have proven a theorem $T$ in some axiom system. How do I show that $T$ does not lead to a contradiction, given that another statement, $T'$, that looks superficially similar, does lead to a contradiction.
If you want us to check your proof, which was perhaps your goal, given that you link to your proof, we'd have to know what axioms you are using. That said, the theorem you say you have proven follows pretty much directly from most axioms for set theory. Given that you haven't asked us explicitly to verify your proof, and that is a lot of work, I will not be attempting that.
Given that you have proven the theorem in some axiom system, it technically "leads to a contradiction" if and only if the original axiom system leads to a contradiction. Given your comments, you are not asking about the consistency of the original axioms, so I will skip this.
Perhaps you mean, "How do I prove this does not directly lead to a contradiction?" The only way to do that is to define "directly," which is actually fairly hard, and is impossible without some explicit axioms.
Finally, perhaps you mean "Why does $T'$ lead to a contradiction, but $T$ doesn't?" This is not a formal question, but it could be a request for clarification of the differences between the two statements that causes on to yield a contradiction but not the other.
So let's write $T'$:
$$\exists R: \forall a: a\in R \iff a\notin a$$
Why does this yield a contradiction? Because if such an $R$ exists, we can ask "Is $R\in R$?" and we get the usual contradiction: $R\in R \iff R\notin R$.
However, your statement, $T$, does not have such a paradox, because it can be rephrased as:
$$\exists R:R\notin R \land R\notin U \land (\forall a: a\in R \iff (a\in U \land a\notin a))$$
Where $U$ is some set. But the reason there is no contraction is the simple addition of that condition on $R$: $a\in U$. There is no contradiction because the requirement "$a\in U$" gives you an "out." That "out" is why the $T$ does not lead to a contradiction, but $T'$ does. It is the difference between $a\notin a$ and $a\notin a \land a\in U$. Once you have that condition, $a\in U$, you lose the contradiction.
Note that $R$ depends on $U$. An $R$ for one $U$ is not equal to the $R$ for another $U$. There is not one universal set $R$ which applies to every $U$ - the conditions, in particular, make it clear that $R\subset U$. (This was why I thought you might be confusing the difference between $\forall\exists$ nd $\exists\forall$ in my early answer - a common mistake.)
As I mentioned in the comments, you re-gain the contradiction if you assert the existence of a universal set: $\exists V:\forall a: a\in V$. But that just means that, in this axiom system, that assertion is invalid - you have essentially proven that the universal set does not exist in your axiom system (or your axiom system is inconsistent.)
But again, it is not clear what you are asking. You haven't asked a well-formed question, and most of the answers and comments posted here have been based on efforts to deduce your meaning based on common types of confusion that plague newcomers to set theory and logic. Are you looking for some formal argument? Are you looking just for clarification/elucidation? Are you looking for a review of your proof? Or are you reaching for something else that none of us have quite grasped yet?