If a group is the union of two subgroups, is one subgroup the group itself?

"Let $G$ be a group, and suppose $G=H \cup K$, where $H$ and $K$ are subgroups. Show that either $H=G$ or $K=G$."

Let $h \in H$ and $k \in K$. Then $hk \in H$ or $hk \in K$ (since every element of $G$ is in either $H$ or $K$). If $hk=h'$ for some $h' \in H$, then $k=h^{-1}h'$, so $k \in H$. If $hk=k'$ for some $k' \in K$, then $h=k'k^{-1}$ so that $h \in K$.

If for all $h \in H$ we have $h \in K$, or if for all $k \in K$ we have $k \in H$, then $H \subseteq K$ or $K \subseteq H$. Then since $G=H \cup K$, we must have either $H=G$ or $K=G$.

I'm not sure if the first paragraph of my 'proof' implies the second. I've shown that for arbitrary $h \in H$, $h \in H$ and possibly $h \in K$, and similar for $k \in K$. I don't know how to wrap it up (or perhaps this route won't lead anywhere at all).

If this way won't work, I'd just like a hint on a new direction to take.

Thanks.

Suppose both $H,K$ are distinct and proper in $G$. Then pick $h\in H\setminus K$ and $k\in K\setminus H$.

In which of $K$ or $H$ or both does $hk$ lie?

Bonus information: as you noted, a group cannot be the union of two of its proper subgroups. It might be interesting to know, that research has been done on how this might generalize.

Theorem (Bruckheimer, Bryan and Muir) A group is the union of three proper subgroups if and only if it has a quotient isomorphic to $C_2 \times C_2$.

The proof appeared in the American Math. Monthly $77$, no. $1 (1970)$. The theorem seems to be proved earlier by the Italian mathematician Gaetano Scorza, I gruppi che possono pensarsi come somma di tre loro sottogruppi, Boll. Un. Mat. Ital. $5 (1926), 216-218$.

For 4, 5 or 6 subgroups a similar theorem is true and the Klein 4-group is for each of the cases replaced by some finite set of groups. For 7 subgroups however, it is not true: no group can be written as a union of 7 of its proper subgroups. This was proved by Tomkinson in 1997.

There is a nice overview paper by Mira Bhargava, Groups as unions of subgroups, The American Mathematical Monthly, $116$, no. $5, (2009)$.