Let $A, B$ be subspaces. I want to show the following: $A \subseteq B \implies B^{\perp} \subseteq A^{\perp}$
Is the following a legitimate proof?
If $x \in A$, then $x \in B$ since $A \subseteq B$. Assume $x$ is not the zero vector. Then because $x \in A$, it must be the case that $x \notin A^{\perp}$. Similarly because $x \in B$ that means $x \notin B^{\perp}$. So we have shown that $x \notin A^{\perp} \implies x \notin B^{\perp}$.
Now by the contrapositive, we see that $x \in B^{\perp} \implies x \in A^{\perp}$. So $B^{\perp} \subseteq A^{\perp}$