I know this statement is true when $G$ is not abelian. But I'm struggling to find a counterexample. I've only considered $S_{4}$ with subgroups $D_{4}$ and $A_{4}$. I'm at a very basic (less than a month old) level of abstract algebra, so I don't know of any other groups with two subgroups.
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Let $G = S_3$ and consider $H = \langle(12)\rangle$ and $K=\langle(23)\rangle$. Then $$|HK| = \frac{|H||K|}{|H \cap K|} = \frac{2\cdot 2}{1} = 4,$$ hence $|HK|$ does not divide $|G|=6$, so by Lagrange's theorem, $HK$ can't be a subgroup of $G$.
In general, $HK$ is a subgroup if and only if $HK = KH$, which is obviously satisfied if $G$ is abelian. A sufficient but not necessary condition is that either $H$ or $K$ is a normal subgroup of $G$.
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$A_4\triangleleft S_4\implies D_4A_4$ is a subgroup. In general, $N$ normal implies $HN$ is a subgroup. So look for subgroups that aren't normal.
How about in $D_4$? Take $H=\{e,\sigma\}$ and $K=\{e,\rho^2\sigma\}$, where $\sigma$ is a reflection and $\rho$ a rotation by $\pi/2$.