The following theorem is an exercise in the section of a textbook dealing with the group isomorphism theorems. However, I did not use the isomorphism theorems to prove this theorem, so I wonder if there might be a cleaner way using them?
Theorem: Let $(A, \cdot)$, $(B, \ast)$ be groups, let $f: A \mapsto B$ be a homomorphism, and let $S^\ast \leqslant B$. Then: $$\textrm{Ker}(f) \leqslant f^{-1}(S^\ast) = \{x \in A \mid f(x) \in S^\ast\} \leqslant A $$
Proof. Let $x, y \in f^{-1}(S^\ast)$. Then, we know that there exists elements $f(x), f(y) \in S^\ast$. Since $S^\ast \leqslant B$, $f(x) \ast f(y) \in S^\ast$. Since $f$ is a homomorphism, $f(x) \ast f(y) = f(x \cdot y) \in S^\ast$. This means that $x \cdot y \in f^{-1}(S^\ast)$, so $f^{-1}(S^\ast)$ is closed under the group operation.
Since $S^\ast$ is a subgroup, we know that if $f(x) \in S^\ast$ (where $x \in f^{-1}(S^\ast)$), then there exists an element $f(y) \in S^\ast$ (where $y \in f^{-1}(S^\ast)$), such that $f(x) \ast f(y) = \textrm{id}_{B}$. Since $f$ is a homomorphism, it must be that $y = x^{-1}$. Thus, $x^{-1} \in f^{-1}(S^\ast)$.
Thus, we have shown that $f^{-1}(S^\ast)$, and every element in it has its inverse within the set too, so $f^{-1}(S^\ast)$ is a subgroup of $A$. It remains to show that $\textrm{Ker}(f) \subset f^{-1}(S^\ast)$. Since $S^\ast$ is a group, $\textrm{id}_{B} \in S^\ast$. Therefore, $\textrm{Ker}(f) = f^{-1}(\textrm{id}_{B}) \subset f^{-1}(S^\ast)$.
I don't believe the isomorphism theorems are necessary in order to prove the theorem. Your proof is the one I would use. Just because the exercise happens to appear in a section dealing with isomorphism theorems does not mean that isomorphism theorems are necessary for the proof. Now, if the exercise had specifically mentioned to use the isomorphism theorems in the proof, that would be different, but again I don't see how the isomorphism theorems would be helpful here.