@Joel David Hamkins has directly answered the first part of my question by providing a simple counterexample which shows that $\mathbf{Cat}$ does not have the Cantor-Schroder-Bernstein property. I will now clarify how to approach the second part of my question.

$\mathbf{Pos}$ usually denotes the category of partially ordered sets as objects and monotone functions as morphisms. For any partially ordered set $(A, \leq)$, we may consider its associated categorification $\mathcal{C}(A, \leq)$ (this is simply viewing $(A, \leq)$ as a category).

There is a natural functor $U: \mathbf{Pos} \rightarrow \mathbf{Cat}$ defined such that, for every partially ordered set $X$, $U(X)=\mathcal{C}(X)$  and such that, for every morphism $f$, $U(f)$ is the unique functor for which the object component of $U(f)$ (in notation $U(f)_{\mathrm{ob}}$) equals $f$.

Now, $U$ is a faithful functor. Thus, to create a counterexample, we only need to show that there is no bimorphism from $(\mathbb{Q}_{\geq 0}, \leq )$ to $(\mathbb{Q}, \leq )$. But this was done by @Peter LeFanu Lumsdaine in [the comments](https://mathoverflow.net/questions/452811/does-mathbfcat-have-the-cantor-schroder-bernstein-property#comment1171201_452814).