Direct proof of empty set being subset of every set Recently I finished my first pure mathematics course but with some intrigue about some proofs of definitions by contradiction and contrapositive but not direct proofs (the existence of infinite primes for example), I think most of them because the direct proof extends away from a first mathematics course or the proofs by contradiction/contrapositive are more didactic. The one that most bothers me in particular is the demonstration that the empty set is a subset of every set, and it is unique. I understand the uniqueness and understand the proof by contradiction:

"Suppose $\emptyset \subsetneq A$ where $A$ is a set. So it exists an element $x \in \emptyset$ such that $x \notin A$ wich is absurd because $\emptyset$ does not have any elements by definition."

but I would like to know if there exists a direct proof of this and if indeed extends from a first course. Thanks beforehand.
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!)
A set $A$ is a subset of a set $B$ if $A$ has no elements that are not also in $B:$

$¬∃x∈A:x∉B$

Since the $empty$ $set$ has $no$ $elements$, it is clearly a subset of every set.