$(X \supset A)\wedge (X \supset B)|(Y\supset A) \wedge (Y \supset B)\to Y\supset X$|Prove that $X=A \cup B$. I've just got this question from Elon Lages' Curso de Análise Vol 1.

Given the sets $A$ and $B$, let $X$ be a set with the following properties:
$1.$ $(X \supset A)\wedge (X \supset B)$.
$2.$ $(Y\supset A) \wedge (Y \supset B)\to Y\supset X$
Prove that $X=A \cup B$.

Reading the property $1$, I know that $A$ and $B$ are in $X$ but it doesn't give the certainty that only $A$ and $B$ are in $X$. When reading $2$, it states that if $Y$ contains both $A$ and $B$, it also contains $X$, this property gives me the certainty that $X$ contains only $A$ and $B$, hence $A \cup B =Y=X$.
Now $A$ and $B$ could be disjoint, they could have an intersection but $A\neq B$ and it could be that $A=B$. Then the formula that would encompass all these possiblities would be $X=A \cup B$.
Would this work as a proof?

I think you've got the right idea, but try to formalize your argument/proof a bit more. You needn't consider all possible cases: that is, 


*

*I. first show that it must follow from property $(1)$ that $A\cup B
   \subseteq X$.

*II. Secondly, show that it must follow from property $(2)$ that $X \subseteq A \cup B$.
Now, having shown that I and II hold, you can conclude that $X = A\cup B$, knowing that for two sets, $X, Y$,  $$(X \subseteq Y \land Y \subseteq X) \iff X = Y$$
