Direct proof of empty set being subset of every set

For any $B\subseteq A$, we have $A\setminus B\subseteq A$. So $A\setminus A=\varnothing\subseteq A$.

(This proof operates at a slightly higher level of abstraction than verifying the definition of $\varnothing\subseteq A$. Since the definition is so easy to verify, you might think that it's silly to take a different strategy. If so, fair enough!)

There is a direct proof, if you know what a vacuous truth is. But the problem is that when one sees this statement $\varnothing\subseteq A$, it's usually before fully understanding vacuous arguments. So it's slightly more instructive to first give a proof by contradiction, and then discuss vacuous arguments. At least from my experience teaching this argument.

The proof is simple. We verify that $\forall x\in\varnothing$ it holds that $x\in A$. However, since $\forall x(x\notin\varnothing)$, the argument holds vacuously. And we are done.

Yes, there's a direct proof:

The way that we show that a set $A$ is a subset of a set $B$, i.e. $A \subseteq B$, is that we show that all of the elements of $A$ are also in $B$, i.e. $\forall a \in A, a\in B$.

So we want to show that $\emptyset \subseteq A$. So consider all the elements of the empty set. There are none. Therefore, the statement that they are in $A$ is vacuously true: $\forall x \in \emptyset, x \in A$. So $\emptyset \subseteq A$.