Your counterexample is correct; indeed it is the universal one, every other counterexample comes from a functor defined on your category. For a counterexample in the category of fields, see my answer here Counterexamples in algebra?.

You seem to be worried about subobjects. If $X$ is an object and $U,V \leq X$ are subobjects such that $U \leq V$ and $V \leq U$, then $U = V$. The reason is that the morphisms $U \to V$ and $V \to U$ over $X$ are uniquely determined (since $V \to X, U \to X$ are monomorphisms). Likewise are the compositions $U \to V \to U, V \to U \to V$ uniquely determined, namely the identity. Thus $U = V$. Thus you don't get into trouble.

Your counterexample is correct; indeed it is the universal one, every other counterexample comes from a functor defined on your category. For a counterexample in the category of fields, see my answer here Counterexamples in algebra?.

You seem to be worried about subobjects. If $X$ is an object and $U,V \leq X$ are subobjects such that $U \leq V$ and $V \leq U$, then $U = V$. The reason is that the morphisms $U \to V$ and $V \to U$ over $X$ are uniquely determined (since $V \to X, U \to X$ are monomorphisms). Likewise are the compositions $U \to V \to U, V \to U \to V$ uniquely determined, namely the identity. Thus $U = V$. Thus you don't get into trouble.