This is simply a summary that includes all the details.

$\mathbf{Pos}$ 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 $U(f)$ equals $f$.


@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. 

In [his answer](https://mathoverflow.net/q/452814 ), Joel considered $(\mathbb{Q}_{\geq 0}, \leq )$ and $(\mathbb{Q}, \leq )$. Given two morphisms $f_1 : (\mathbb{Q}_{\geq 0}, \leq ) \rightarrow (\mathbb{Q}, \leq )$ and $f_2: (\mathbb{Q}, \leq ) \rightarrow (\mathbb{Q}_{\geq 0}, \leq )$ in $\mathbf{Pos}$ (monotone maps), $U(f_1): \mathcal{C}(\mathbb{Q}_{\geq 0}, \leq ) \rightarrow \mathcal{C}(\mathbb{Q}, \leq )$ and $U(f_2): \mathcal{C}(\mathbb{Q}, \leq ) \rightarrow \mathcal{C}(\mathbb{Q}_{\geq 0}, \leq )$ are easily seen to be monomorphisms in $\mathbf{Cat}$ (these functors are injective on objects since any monotone map is injective and are faithful vacuously). However, it doesn't hold that $\mathcal{C}(\mathbb{Q}_{\geq 0}, \leq ) \cong \mathcal{C}(\mathbb{Q}, \leq )$ since $\mathcal{C}(\mathbb{Q}_{\geq 0}, \leq )$ has an initial object and $\mathcal{C}(\mathbb{Q}, \leq )$ does not.

As was mentioned by @Peter LeFanu Lumsdaine in [the comments](https://mathoverflow.net/questions/452811/does-mathbfcat-have-the-cantor-schroder-bernstein-property#comment1171201_452814), it turns out that Joel's counterexample can be extended to completely answer my question. Namely, $U$ is a faithful functor. Thus, to show that there are no bimorphisms from $\mathcal{C}(\mathbb{Q}_{\geq 0}, \leq )$ to $\mathcal{C}(\mathbb{Q}, \leq )$ in $\mathbf{Cat}$, we only need to show that there is no bimorphism in $\mathbf{Pos}$ from $(\mathbb{Q}_{\geq 0}, \leq )$ to $(\mathbb{Q}, \leq )$.

Peter defined, for every morphism $g : (\mathbb{Q}_{\geq 0}, \leq ) \rightarrow (\mathbb{Q}, \leq )$ in $\mathbf{Pos}$, a morphism $\tilde{g} : (\mathbb{Q}, \leq ) \rightarrow (\mathbb{Q}, \leq )$ defined by the rule

\begin{equation*}
\forall x \in \mathbb{Q}:  \tilde{g}(x)=\left\{\begin{matrix}
x, & x \geq g(0)\\ 
x-1, & x < g(0)
\end{matrix}\right. .
\end{equation*}

It holds that $\tilde{g} \circ g=g$. Thus, $g$ is not an epimorphism and is thus not a bimorphism. We conclude that there exists no bimorphism in $\mathbf{Pos}$ from $(\mathbb{Q}_{\geq 0}, \leq )$ to $(\mathbb{Q}, \leq )$.