Proving set identities: empty set case.

In order to show that for two sets $X$ and $Y$ it holds that $X\subseteq Y$, you have to prove that

for every $x$, if $x\in X$, then $x\in Y$.

Note that a statement of the form “if $\mathscr{A}$ then $\mathscr{B}$” is true when

either $\mathscr{A}$ is false or both $\mathscr{A}$ and $\mathscr{B}$ are true

If $X$ is the empty set, then “$x\in X$” is false for every $x$; hence “if $x\in X$ then $x\in Y$” is true.

The phrase “take an arbitrary element $x\in X$” is possibly misleading, but its intended meaning is “suppose $x\in X$”.

If $D$ is the empty set then the following statement is always true:$$\text{If }x\in D\text{ then }\cdots$$no matter what the dots are standing for.

It is false that $x\in D$ and "ex falso sequitur quodlibet". A false statement implies whatever you want.

So the assumption will not bring you into troubles, but - on the contrary - will give you freedom to accept whatever you want.

Since the empty set is a subset of any set, there is no need of including that in a formal proof. However it is a good idea to be aware of the fact that the equality holds even in the case of empty sets .