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.

Here is another direct proof, more calculational, where we first use the definitions and basic properties of $\;\emptyset,\subseteq\;$ and then simplify using predicate logic:
\begin{align}
& \emptyset \subseteq A \\
\equiv & \qquad\text{"definition of $\;\subseteq\;$"} \\
& \langle \forall x :: x \in \emptyset \Rightarrow x \in A \rangle \\
\equiv & \qquad\text{"basic property of $\;\emptyset\;$"} \\
& \langle \forall x :: \text{false} \Rightarrow x \in A \rangle \\
(*) \quad \equiv & \qquad\text{"logic: false implies anything"} \\
& \langle \forall x :: \text{true} \rangle \\
\equiv & \qquad\text{"logic: leave out unused quantified variable"} \\
& \text{true} \\
\end{align}
Note how the last steps are really just a more detailed proof of the principle of 'vacuous truth', as used by earlier answers.
If you want more detail on the third step $(*)$:
\begin{align}
& \text{false} \Rightarrow P \\
\equiv & \qquad\text{"rewrite"} \\
& \lnot \text{false} \lor P \\
\equiv & \qquad\text{"simplify"} \\
& \text{true} \lor P \\
\equiv & \qquad\text{"simplify"} \\
& \text{true} \\
\end{align}
