Prove the empty set is a subset of every set. Does induction work? 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?
 A: 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. 
A: Induction works for the natural numbers. Namely, you've shown that for every finite set, the empty set is a subset of that finite set.
But traditionally set theory is concerned about infinite sets as well. So your proof will not cover those. You can complete the proof by showing that every non-empty set has a finite subset, in which case the proof is complete. 
But then why not show that every non-empty set has a singleton subset and only prove that the empty set is a subset of singletons? But since we reached singletons, why not just show using a vacuous truth argument that the empty set is a subset of every set?
