Prove that the empty set is a subset of every set. The exercise is taken from Rudin, principles of mathematical analysis, chapter 2 ex. 1.
Let $A$ a set and let also $B$ such that $A \cap B = \emptyset$ This implies:
$$
\emptyset = A \cap B \subseteq A,
$$
For given $A$ the set $B$ can always be found, for example take $B = A^C$, the complement of $A$. Is such proof correct?
 A: Looks good. We can also do it like this:
$A \subset B$ is equivalent to saying $x \in A \implies x \in B$. This statement is vacuously true when $A = \varnothing$ (since there is no $x \in A$) and $B$ is any set. (even the empty set!)
A: You're missing the essential point. In fact, I'd say this is essentially circular. How do you know that the intersection of two sets is necessarily a subset of both of them? If $A \cap B$ is non-empty, then let $x \in A \cap B$  and, by definition, $x \in A \land x \in B$ which proves the theorem. However,  if $A$ and $B$ are disjoint, then this proof isn't quite as obvious, is it? 
To prove the exercise, note that $x \in \emptyset  \implies x \in S$ means the same thing as $x \notin \emptyset \ \lor x \in S $. However, $\forall x, x \notin \emptyset$ and so the theorem is proved. 
A: $$\emptyset \subset A \equiv (\forall x) (x \in \emptyset \Rightarrow x \in A) \equiv \neg (\exists x) (x \in \emptyset \land x \notin A)$$
As $\emptyset$ is the empty set, $x \in \emptyset$ can never be true, i.e., $x \in \emptyset \land x \notin A$ is always false. Thus, $(\exists x) (x \in \emptyset \land x \notin A)$ is false and its negation is true. Hence, $\emptyset \subset A$ is true regardless of $A$.
A: I don't like it.
It works in the context of some set $X$ serving as universe and containing $A$ as subset. 
In my view preferable is: $\varnothing=A-A\subseteq A$.
Even more preferable: you can just prove that every element of $\varnothing$ is an element of $A$. This is vacuously true: no element of $\varnothing$ can be found that is not an element of $A$. Even stronger: no element of $\varnothing$ can be found at all.
