Can this be done by induction instead of just proving the subgroup criterion? I can prove using the essentials tools of group theory, but looking at the problem, I was wondering if we can simply use an induction argument.

Given we have a chain of ascending subgroups of a group $G$... $$H_1\le H_2\le....$$ Is the union $$\bigcup_{i=1}^\infty{H_i}\le G$$

For the first case, we have that $H_1\le H_2$ and thus, $H_1\subseteq H_2$ means that in terms of their union,

$$\bigcup_{i=1}^2{H_i}=H_2\le G$$

Thus, for an arbitrary $n>2$ we can assume that $$\bigcup_{i=1}^n{H_i}=H_n\le G$$ Then $$\bigcup_{i=1}^{n+1}{H_i}=\left(\bigcup_{i=1}^{n}{H_i}\right)\cup H_{n+1}=H_n\cup H_{n+1}=H_{n+1}\le G$$

Thus, for any whole value of $n$, the claim is true.

My worry is the infinite upper bound. But I feel like induction takes care of that.