In a topos, retracts of objects of the form $\Omega^Y$ are injective I am trying to prove the statement from the title. Let $X$ be a retract of $\Omega^Y$, say $r: \Omega^Y \to X$ with right inverse $s$. 
Suppose we have a diagram $B \xleftarrow{f} A \xrightarrow{m} X$ with $m$ mono. If $m \circ r$ is mono, then we are done because we can use that $\Omega^Y$ is an injective object, but this is not true in Set, so this approach fails. 
I can't think of another way to approach this problem.
 A: Side comment:
You have the wrong map being a monomorphism to verify injectivity, it should be $f$ in your set up if you want to check that $X$ is injective.

Main Answer:
This is true in more generality, if you already know $\Omega^Y$ is injective.

In any category $C$, retracts of injectives are injective.

Proof:
Assume $Y$ is injective, and let $X$ be a retract:
$$X\newcommand\toby\xrightarrow\toby{i} Y\toby{r} X, \quad ri=\newcommand\id{\operatorname{id}}\id_X.$$
Now suppose we have a diagram 
$$
\require{AMScd}
\begin{CD}
A @>j>> B \\
@VfVV @.\\
X,
\end{CD}
$$
with $j$ a monomorphism. Then
augment the diagram by mapping $X$ to $Y$ by $i$.
$$
\begin{CD}
A @>j>> B \\
@VfVV @.\\
X @>i>> Y.
\end{CD}
$$
Now since $Y$ is injective, there exists some map $g:B\to Y$ making the diagram commute: 
$$
\begin{CD}
A @>j>> B \\
@VfVV @VVgV\\
X @>i>> Y.
\end{CD}
$$
Finally, the map $rg : B\to X$ is the desired map making the original triangle commute, since 
$rgj = rif=f$. $\blacksquare$

Additional side comments:
This proof immediately generalizes to retracts of $\mathcal{H}$-injectives are injective for any class of morphisms $\mathcal{H}$, and since the dual of a retract is a retract, the dual theorem is that retracts of ($\mathcal{H}$-)projectives are ($\mathcal{H}$-)projective.
