Category of Subobjects is Category
Theorem
Let $\mathbf C$ be a metacategory.
Let $C$ be an object of $\mathbf C$.
Let $\map {\mathbf{Sub}_{\mathbf C}} C$ be the category of subobjects of $C$.
Then $\map {\mathbf{Sub}_{\mathbf C}} C$ is a metacategory.
Proof
Let us verify the axioms $(C1)$ up to $(C3)$ for a metacategory.
Let $f: m_1 \to m_2$ and $g: m_2 \to m_3$ be morphisms of $\map {\mathbf{Sub}_{\mathbf C}} C$.
That $g \circ f: m_1 \to m_3$ is again a morphism follows from the following commutative diagram in $\mathbf C$:
$\quad\quad \begin{xy}\xymatrix@+2em{ \operatorname{dom} m_1 \ar[r]_*+{f} \ar[rd]_*+{m_1} \ar@/^1pc/[rr]^*+{g \circ f} & \operatorname{dom} m_2 \ar[r]_*+{g} \ar[d]^*+{m_2} & \operatorname{dom} m_3 \ar[ld]^*+{m_3} \\ & C }\end{xy}$
For every subobject $m$ of $C$, the diagram:
$\quad\quad \begin{xy}\xymatrix{ \operatorname{dom} m \ar[r]^*+{\operatorname{id}_m} \ar[rd]_*+{m} & \operatorname{dom} m \ar[d]^*+{m} \\ & C }\end{xy}$
in $\mathbf C$ commutes, for $\operatorname{id}_m = \operatorname{id}_{\operatorname{dom} m}$, and the latter is an identity morphism in $\mathbf C$.
This last property also immediately proves the required behaviour of $\operatorname{id}_m$ with respect to other morphisms of $\map {\mathbf{Sub}_{\mathbf C}} C$.
Since the composition is inherited from $\mathbf C$, it is necessarily associative.
Hence $\map {\mathbf{Sub}_{\mathbf C}} C$ is a metacategory.
$\blacksquare$