I've taken a look at the proofs by contrapositive and by vacuous truths (for the above title), but I was wondering whether or not the following proof by induction works.
The following proof proceeds by induction on the number of elements in a given set. Let $\phi$ be some set with one element and $\phi\prime$ be some nonempty set that shares no common elements with $\phi$. Then by definition, $$\phi \cap \phi\prime = \emptyset$$ Which implies that $\emptyset \subseteq\phi$, establishing the base case. Then for the inductive step, let $\psi$ be some set with $n+1$ elements and let $\psi\prime$ be some nonempty set that shares no common elements with $\psi$. Then, $$\psi \cap \psi\prime = \emptyset$$ Meaning that $\emptyset \subseteq\psi$, which proves the inductive step, completing the proof by induction.
Is this logic fine?
No, induction does not work since not every set is finite.
Notice that your proof is essentially the following one. Using the fact that $A\cap B\subseteq A$ for all sets $A,B$ (a fact you need to prove), given any set $A$, let $B=\{b\}$ with $b\notin A$ (you need to show such a $b$ exists). Then $A\cap B=\emptyset$ and therefore $\emptyset \subseteq A$. This does not require induction at all.