I mostly just want to double-check my reasoning in my proof. For clarity's sake, the diagram we are working with is
where $m$ is monic, the top-left corner is a pullback, and we wish to show that $m'$ is monic.
Here's a sketch of my proof. I can provide more thorough reasoning if wanted, I'm just providing the rough outline of my reasoning for brevity's sake. (I was able to prove each individual thing well enough to satisfy myself, I just want to be sure the overall reasoning itself is all right.)
First, we begin by considering another object $X$ with parallel arrows $a,b : X \rightarrow M'$ such that $m' \circ a = m' \circ b$
Before even beginning the proof proper, we note the end goal: to prove $m'$ is monic. Our construction helps with that, as will be seen. The goal is to show $a=b$ when $m' \circ a = m' \circ b$. The goal, then, is to invoke the universal property of the pullback: in doing so, we assure there exists a unique arrow $u : X \rightarrow M'$. Because of this, it follows $a = b = u$. And thus $m'$ is monic, because our assumption $m' \circ a = m' \circ b$ led to $a = b$.
Through our assumptions from the construction of $X, a, b$ and from the facts that $m$ is monic and the original diagram is a pullback, we next need to prove the commuting of the diagram when X is in play. A rough look at it:
It's not pretty to look at - I should really learn how to do arrows in LaTeX - so for clarification's sake, the essential equalities we need to verify are:
$$g\circ a=g\circ b\\ m'\circ a=m'\circ b\\ m\circ (g\circ a)=f\circ (m'\circ a)\\ m\circ (g\circ b)=f\circ (m'\circ b)$$
The first is the commuting of the top trapezoid, the second of the left one, and the third and fourth assure the commuting of the outer square.
With the necessary commutings assured, we can invoke the universal property of the pullback. Thus, $\exists ! u : X \rightarrow M'$, etc. etc.
Since $u$ is unique, $a = b = u$. Since our assumption $m' \circ a = m' \circ b$ implied $a = b$, alongside the other assumptions (pullback, $m$ monic) then $m'$ is monic.
For what it's worth, I believe this is essentially the same line of reasoning communicated in this question. I'm mostly just elaborating further on it, sort of explaining it to myself if you will, since it didn't 100% make sense to me when I saw it there.