Let T be a topology on a set X such that T consists of precisely four sets; that is , $T = \{X, \emptyset, A, B\}$, where $A$ and $B$ are non empty distinct proper subsets of $X$. Prove that $A$ and $B$ must satisfy exactly one of the following conditions: 

a) $B = X\setminus A$  

b) $A \subset B$  

c) $B \subset A$

hint: show that A and B must satisfy at least one of the conditions and thens show they cannot satisfy more than one of the conditions.

Attempt: So I tried this, but it doesn't seem to complete in my opinion.

Assume a) i.e $B = X\setminus A$:  

Then $A\cup B = X \in T$ and $A \cap B = \emptyset \in T$

So I established that it satisfies the topological conditions, but now to show that the other conditions cannot be satisfied at the same time how could I go about it? I mean if I explain it in just words it would be that since I assumed condition a) then there is no way for b) or c) to even be valid. But there has to be a way to show it no?