$|A|$ is the cardinality of $A$ which is defined to be the least ordinal such that there exists bijective map between $A$ and $|A|$
There exists an injective map but no bijective map from $A$ to $B$ $\implies |A|<|B|$($<$ is the ordering between ordinals by $\in$ or equivalently $\subsetneq$).
Proof) Since there doesn't exist a bijective map from $A$ to $B$, $|A|\neq|B|$. And since $|A|,|B|$ are ordinals, either $|A|<|B|$ or $|A|>|B|$. Suppose $|A|>|B|$. Then there exists an injective map from $B$ to $A$. Then by Cantor-Bernstein theorem, there exists a bijective map from $A$ to $B$ and $|A|=|B|$, a contradiction. So $|A|<|B|$.
I was wondering if this is a correct proof.