If $Ha\subseteq Kb$ for some $a,b\in G$, show that $H \subseteq K$. 
Let $H$ and $K$ be subgroups of a group $G$. If $Ha\subseteq Kb$ for some $a,b\in G$, show that $H \subseteq K$.

I constructed a proof by contradiction and I am wondering whether or not it is free from flaws. I thank you in advance for taking your time to inspect my work!
Let us assume to the contrary that $H \nsubseteq K$. Then $\exists h \in H$ such that $h \notin K$. As $Ha\subseteq Kb$, then $H \subseteq Kba^{-1}$. So for this $h \in H$, it is of the form $h=kba^{-1}$ for some $k \in K$. 
Hence, $h=kba^{-1} \notin K$ by hypothesis $\Rightarrow h=kba^{-1} \in Kc$ for some $c \in G$, which then implies that $k \in Kcab^{-1}$. This last implication only holds if $cab^{-1}=1$ $\Rightarrow c=ba^{-1}$.
As $h=kba^{-1} \in Kc$, then $h=kc \in Kc \Rightarrow h \in K$, which contradicts our assumption and thus concludes the proof. 

Put $g=ba^{-1}$, then $H \subseteq Kg$. Since $1 \in H$, we have $1=kg$ for some $k \in K$. Hence $g=k^{-1}$ and $Kg=Kk^{-1}=$ (since $K$ is a subgroup )$K$, so $H \subseteq K$.
