Consider $A$ and $B$ subobjects of $C$ in an abelian category, and $\iota_A : A \rightarrow C$ and $\iota_B : B \rightarrow C$ the inclusions as subobjects.
On the one hand we have $a : A\rightarrow A\cup B$ and $b : B\rightarrow A\cup B$ the inclusion monomorphisms from the definition of union as the join of subobjects.
Using matrix notation we may have $(a\ b) : A\oplus B \rightarrow A\cup B$ and $(\iota_A\ \iota_B) : A\oplus B \rightarrow C$.
On the other hand we have the union as the epi-mono factorization of $(\iota_A\ \iota_B)$ as $\require{AMScd}$ \begin{CD} A\oplus B @>{coim\ (\iota_A\ \iota_B)}>> A\cup B @>{im\ (\iota_A\ \iota_B)}>> C \end{CD}
Is it the case that $coim\ (\iota_A\ \iota_B) = (a\ b)$ ?
Recall that the epi-mono factorization is only unique up to isomorphism, so the best we can do is show that $(a\ b) : A \oplus B \to A \cup B$ followed by the inclusion $A \cup B \hookrightarrow C$ is an epi-mono factorization of $(\iota_A\ \iota_B)$.
Consider the following diagram :
We defined $(\iota_A\ \iota_B)$ to be the factorization of the cocone given by $\iota_A$ and $\iota_B$, but $a$ and $b$ each followed by the inclusion $A \cup B \hookrightarrow C$ give the same cocone, hence $(\iota_A\ \iota_B)$ indeed factors through $(a\ b)$ : $$(\iota_A\ \iota_B) : A \oplus B \xrightarrow{(a\ b)} A \cup B \hookrightarrow C$$ The second morphism is clearly a monomorphism, hence it remains to show that $(a\ b)$ is epic.
Since we are working in a (small) abelian category $\mathbf{C}$, we can use the Freyd-Mitchell embedding theorem : $\mathbf{C}$ is equivalent to some full subcategory of some module category $R-\mathrm{Mod}$. In such a category of modules over a ring, $A \oplus B \to A \cup B$ is given by $(a, b) \mapsto a+b$ (as $A \cup B$ is none other than $A + B$), which is surjective, thus an epimorphism. Now epimorphisms in $R-\mathrm{Mod}$ are epimorphisms in the subcategory $\mathbf{C}$, so $(a\ b)$ is indeed epic in $\mathbf{C}$.
Maybe (I hope) there is a nicer argument that doesn't use Freyd-Mitchell.