Let $\bar{\bar{x}}$ denote the cardinal of $x$ and $\approx$ denote bijective equivalence.
Assume $x\preceq y$. By definition $\exists z (z \subseteq y \land x \approx z)$. Now from something I've already proven as $x \approx z$ then $\bar{\bar{x}} = \bar{\bar{z}}$. Thus all I need to prove is that $z \subseteq y \to \bar{\bar{z}} \leq \bar{\bar{y}}$. As cardinals are just ordinals I've attempted to prove this by contradiction. I.e assume $ \bar{\bar{y}} < \bar{\bar{z}}$, so $ \bar{\bar{y}} \in \bar{\bar{z}}$. Now as $z \approx \bar{\bar{z}}$ and $y \approx \bar{\bar{y}}$ I think that our assumption should lead to a contradiction of the fact that $z \subseteq y$.
I don't know why but after a few pages of scratchings I'm not able to clearly reach the desired contradiction. This seems like it should be a simple proof so I don't know if I'm going down the wrong path or if I'm missing something glaringly obvious. I was able to prove the converse in a few lines. Some guidance would be much appreciated.