How to prove X = A $\cup$ B given subset conditions? Problem
Given sets $\mathcal A$, $\mathcal B$, $\mathcal Y$, let $\mathcal X$ be a set with the following properties:


*

*$\mathcal X$ $\supset$ $\mathcal A$ and $\mathcal X$ $\supset$ $\mathcal B$,

*if $\mathcal Y$ $\supset$ $\mathcal A$ and $\mathcal Y$ $\supset$ $\mathcal B$  , then $\mathcal Y$ $\supset$ $\mathcal X$
Prove that $\mathcal X$ = $\mathcal A$ $\cup$ $\mathcal B$.

My work
Let $\mathcal x$ be an element of set $\mathcal X$ and $\mathcal y$ be an element of set $\mathcal Y$.
If $\mathcal X \supset\mathcal A$, then $\mathcal x$ $\in$ $\mathcal A$, then $\mathcal x$ $\in$ $\mathcal X$. 
 If $\mathcal X \supset\mathcal B$, then $\mathcal y$ $\in$ $\mathcal B$, then $\mathcal y$ $\in$ $\mathcal Y$.
If $\forall$ $\ x$ $\in$ $\mathcal A$, $\ x$ $\in$ $\mathcal X$; and $\forall$ $\ y$ $\in$ $\mathcal B$, $\ y$ $\in$ $\mathcal Y$, then $\mathcal X = \mathcal A \cup \mathcal B$.

This is an introductory problem in a Real Analysis course I'm doing on my own. I'm an Engineer trying to acquire a deeper mathematical maturity. The book I'm following doesn't bring answers, ergo, the question. But more than a "Right or Wrong" answer, I'd like an evaluation concerning the rigor - or the lack thereof - of the answer I provided. 
 A: Your proof is not correct. There is no $\mathcal Y$ to begin with so your first statement doesn't make sense. 
First verify that $\mathcal A \cup \mathcal B \subset \mathcal X$. This is easy from the first condition and the definition of union.
Next let us prove that $\mathcal X \subset \mathcal A \cup \mathcal B$.  Let us prove this by contradiction. Suppose $x \in \mathcal X \setminus \mathcal A \cup \mathcal B$. Take $\mathcal Y=\mathcal A \cup \mathcal B \setminus \{x\}$. Use the second condition given to show that we must have $\mathcal X \subset\mathcal A \cup \mathcal B \setminus \{x\}$. Arrive at a contardiction by observing that $x$ belongs to LHS but not to RHS. 
A: The thing is that $A,B, X$ are specific sets that exist and a given to you.  They do not change.  
$Y$ can be any set in the world.
So let $Y = A\cup B$. 
(We know in general that $A \subset A\cup B$ and $B\subset A\cup B$.  This isn't just true of these sets.  This would be true for any sets, $K$ and $M$.  It is always true that $K \subset K\cup M$. That is because  $K\cup M$ is the set of all elements that are in $K$ or in $M$.  If $x \in K$ then it is one of the elements that are in $K$ or in $M$.  So all the elements of $K$ are elements in $K\cup M$.  So $K\subset K\cup M$.)
$A\subset A\cup B$ or in other words $Y= A\cup B \supset A$.  And $B\subset A\cup B$ or in other words $Y= A\cup B \supset B$.
But we have a condition that if  $Y\sup A$ and $Y\sup B$ then we will have $Y\sup X$.
And we have do have that $A\cup B \supset A$ and $A\cup B \supset B$ so we must have $A\cup B\supset X$.
On the other hand:  $A\cup B \subset X$.  We know this because if $x \in A\cup B$ then either $x \in A$ or $x\in B$.  If $x \in A$ then, because $X\sup A$ so every element in $A$ is in $X$, we know $x \in X$.  And if $x \in B$ then $x\in X$ because $X\sup B$ and that's what $X\sup B$ means.  So either way $x \in X$.  
So every element in $A\cup B$ is in $X$ so $A\cup B \subset X$.
We also have $A\cup X \supset X$ so every element in $X$ is in $A\cup B$.
So $A\cup B$ and $X$ both have the exact same elements.
So $A\cup B = X$.
.... 
As to your work:
"x be an element of set X... If X⊃A, then x ∈ A,then x ∈ X"
"If $X \subset A$"  No if about it!  You were told that $X \supset A$ so nothing to speculate.
"then $x \in A$".  That's not true.  Just pick an $x$ that is in $X$ but not in $A$.  Then $x$ will not be in $A$.
"then x ∈ X"  But... you already said that.
... and so on.
