Set equivalence - Zorich's real analysis I am doing my BS Mathematics second year through an open university IGNOU.
I am taking a first course in real analysis. I am using the text by Bartle and Shebert and the one by Zorich to supplement it. I would like to have my proof on elementary set equivalence verified. This is a naive attempt.

Verify the relation
$$(A \subset C)\wedge(B \subset C)\iff((A \cup B) \subset C)$$

Proof.
(1) Consider arbitrary elements $x_{1}\in A,x_{2} \in B$.
We are given $(A \subset C) \wedge (B \subset C)$. It follows -
$x_{1} \in A \implies x_{1} \in C$.
$x_{2} \in B \implies x_{2} \in C$.
$x_{1},x_{2} \in (A \cup B) \implies x_{1},x_{2} \in C$
Consequently, $(A \cup B) \subset C$.
$(A \subset C) \wedge (B \subset C) \implies (A \cup B) \subset C$.
(2) In the other direction, consider arbitary elements $x_{1},x_{2} \in (A \cup B)$. Given $(A \cup B) \subset C$, we must have $x_{1},x_{2} \in C$.
But, $x_{1},x_{2} \in (A \cup B)$ implies
(a) $x_{1} \in A, x_{2} \in B$ or
(b) $x_{1},x_{2} \in (A \cap B)$ or
(c) $x_{1},x_{2} \in A, B = \phi$ or
(d) $A = \phi, x_{1},x_{2} \in B$
Each of these implies the left hand side $(A \subset C)\wedge(B \subset C)$ holds.
$((A \cup B) \subset C) \implies (A \subset C)\wedge(B \subset C)$
From (1) and (2) we have :
$((A \cup B) \subset C) \iff (A \subset C)\wedge(B \subset C)$
 A: You don't need two elements, just one arbitary one will do to show an inclusion.
So suppose $(A \subset C) \land (B \subset C)$ holds.
This implies that $A \subset C$ and $B \subset C$ both hold.
We want to show that $(A \cup B) \subset C$, so let $x \in A \cup B$ be arbitary.
This means that $x \in A$ or $x \in B$. If the first is the case, then from  $A \subset C$ we conclude that $x \in C$. If the second is the case, from $B \subset C$ we see that $x \in C$. So $x \in C$ holds. This shows that $(A \cup B) \subset C$.
Now suppose that $(A \cup B) \subset C$ holds, and we want to show that $(A \subset C) \land (B \subset C)$ is then true.
So take $x \in A$ arbitrarily. Then $x \in A \cup B$ and so $x \in C$ by the assumed inclusion. So $A \subset C$.
Take $x \in B$ arbitarily. Then $x \in A \cup B$ and so $x \in C$ by the assumed inclusion. So $B \subset C$. As both hold, so does $(A \subset C) \land (B \subset C)$.
We have shown two implications, hence the equivalence.
A: The $x_1, x_2$ stuff is just confusion.  
Proof of left to right.  Assume the premise.
If x in A $\cup$ B, then x in A or x in B.
In either case, by the premise, x in C. 
Proof of right to left is simple.  All it takes is knowing
A subset A $\cup$ B and B subset A $\cup$ B,
