Proof involving a maximal subgroup

Question

Let $G$ be a group with $H \leq G$. Suppose that $x\notin H$ and define $U = \langle x, H\rangle$. Choose a subgroup $M$ of $U$ which is maximal under the properties $H\leq M$ and $x\notin M$. Then $M$ is a maximal subgroup of $U$.

Keep in mind this is part of a longer proof and the author just says it is clear.

I'm not sure how to proceed with this proof. I want to start by letting $K$ be a subgroup such that $M \leq K \leq U$. I need to show that either $M=K$ or $K=U$.

Answer 1

If $K\ne M$, then $x\in K$, otherwise $M$ is not maximal with respect to the given property, because $K$ also satisfies it. But if $x\in K$ and $H\subseteq K$ then $U\subseteq K$ by definition of $U$. So $K=U$.