Could anyone check if my proof is ok please?
$g\circ f$ is monic if $f$ and $g$ are monic
$g\circ f$ being monic means that for two parallel arrows $i$ and $j$ (ie they share the same domain and codomain), if $g\circ f \circ i=g\circ f \circ j$ then $i=j$. So we begin by assuming $g\circ f \circ i=g\circ f \circ j$.
Here we apply the def. of monic twice: since $g$ is monic, so $f\circ i = f\circ j$. Likewise since $f$ is monic, $i=j$.
Yes, the proof is valid.
Associativity of $\circ$ allows one to omit parentheses but, at this stage, it might help to leave them in to see what's really going on. It's not necessary though.