Let $G$ be a group and $K\le G, H\le G$. If there exist $x,y\in G$ such that $xK\subseteq yH$, then $K\le H$.
Solution 1:
Since $xK\subseteq yH$, $K\subseteq x^{-1}yH$. So, since $e\in K$, $e\in x^{-1}yH$. That is, $y^{-1}x\in H$. So, its inverse, which is $x^{-1}y$, also belongs to $H$, and therefore $x^{-1}yH$ is simply $H$. So, since $K\subseteq x^{-1}yH$, $K\subseteq H$.
Solution 2:
From your argument, you have shown that $y^{-1}xk \in H$ for every $k \in K$.
In particular, it holds for the identity element $e$, which lies in $K$ because $K$ is a subgroup.
Thus we have $y^{-1}x \in H$. It follows that $k = (y^{-1}x)^{-1} \cdot (y^{-1}x) \in H$ for any $k \in K$.