Working through the First Isomorphism Theorem in Pinter's A Book of Abstract Algebra and I want to make sure this proof is not using circular logic via my definitions or notation.
Let $H$ be a normal subgroup of $G$ and let $K$ be any subgroup of $G$. Assume for the purposes of contradiction that $H\cap K$ is not a normal subgroup of $K$. Then there must exist some $k\in K$ such that for some $b\in H\cap K$ that $kbk^{-1}\not\in H\cap K$. But since $H$ is normal, then $kbk^{-1}\in H$ since $H$ must be closed with respect to conjugates when $b\in H$. So $kbk^{-1}\not\in K$ in order for $kbk^{-1}\not\in H\cap K$. But $b$ is also in $K$ by definition of the intersection and $k,k^{-1}\in K$. Since $K$ is a group and must be closed under its operations $kbk^{-1}\in K$ which causes our contradiction. Thus $H\cap K$ must be normal.
Thanks!
Your proof is fine but it'd honestly be cleaned up a bit by just doing it "directly": so for instance, you can just take any $k\in K$ and $h\in H\cap K$, then note that $khk^{-1}\in H$ since $H$ is normal, and that $k,h,k^{-1}$ are all in $K$ so their product $khk^{-1}$ is in $K$. Thus we've shown $khk^{-1}\in H\cap K$.
Note this is basically the exact argument you've written, just without an unnecesary hypothesis at the beginning to end in a contradiction.