Let $C$ be an abelian category, and consider the following diagram
$\require{AMScd}$ \begin{CD} \ker(C\to B)@>h>> C@>>f\circ p>B \\ @. @VpVV @V idVV \\ \ker(A\to B) @>k>>A @>>f> B \end{CD}
Assume that $p$ is an epimorphism and that $h\circ p = 0$, how does one prove $f$ is monic, i.e. $k=0$. If one were able to use elements, it's fairly easy, take $a\in A$ such that $f(a)=0$, then you can lift it to $c\in C$ but then by commutativity of the diagram $f\circ p(c)=0$ hence $c\in \ker(C\to B)$, hence $p (c)=0$ but $p (c) = a$.
The only part that's not straightforward in a general setting is to lift $k$ to $C$, when using elements, surjectivity allows us to do that but I don't know how to replicate that here.
Note that I'm aware I could use for instance the Freyd-Mitchell theorem to be able to carry out the proof with elements and conclude it's true in the general setting as well but this shows up in the proof of Gabriel-Popescu's theorem which is another embedding theorem and I'd like to not use an embedding theorem to prove that, I'm fairly certain it's not necessary.
$\newcommand{\coker}{\operatorname{coker}}$We can use the generalised diagram chase (see MacLane's Categories for the Working Mathematician for a proof of the following rules, or have a go yourself):
Here, a statement like "$x\in_m a$" just means ("$m$" for "member"), there is some object $X$ and some arrow $x:X\to a$. If $x,y\in_m a$ are equivalent ($x\equiv y$) then that means to say: "there are epimorphisms $u,v$ with $xu=yv$". Of course, $\equiv$ is an true equivalence relation on "members" of any object.
Suppose $x\in_m A$ and $fx\equiv 0$. Since $p$ is an epimorphism, by rule $\rm(iii)$ we can say that there exists a $y\in_m C$ with $py\equiv x$. Then, $0\equiv fx\equiv fpy$ so by rule $\rm(v)$ we know there exists some $z\in_m\ker(C\to B)$ with $hz\equiv y$. Then $x\equiv py\equiv phz\equiv0$ (using $\rm(iv)$ on $ph=0$) and by rule $\rm(i)$ we can conclude $f$ is monic.
I suspect that there may be a more rudimentary way to do this with direct manipulation of categorical constructions but hey, the diagram chase is fun and useful to know about.