Powering-invariance is union-closed

Statement
Suppose $$G$$ is a group and $$H_i,i \in I$$ is a collection of powering-invariant subgroups of $$G$$. Suppose the union $$\bigcup_{i \in I} H_i$$ is a subgroup of $$G$$ (and hence is also the same as the join of subgroups $$\langle H_i \rangle_{i \in I}$$). Then, this union of also a powering-invariant subgroup of $$G$$.

Related facts

 * Quotient-powering-invariance is union-closed
 * Powering-invariance is not finite-join-closed

Proof
Given: $$G$$ is a group and $$H_i,i \in I$$ is a collection of powering-invariant subgroups of $$G$$. The union $$H = \bigcup_{i \in I} H_i$$ is a subgroup of $$G$$. $$G$$ is powered over a prime $$p$$. An element $$g \in H$$.

To prove: There exists $$x \in H$$ such that $$x^p = g$$.

Proof: