How to prove that if the set and its complement are recursively enumerable, then both are recursive?
Roughly, yes, that's the right idea.

I think there are a few small issues though.

First, you can not compute the characteristic function $\chi_A$, since this is computable only when the set is recursive. What you can compute is the semi-characteristic function, whose domain is $A$, and is constantly $1$ there.

If we wanted to be more pedantic, "computing a function step by step" is not completely precise. One usually takes a "program" implementing the semi-characteristic (e.g. a Turing machine accepting each element of $A$, and diverging elsewhere) and executes that program step-by-step.

So, given two TMs $M$ and $N$ implementing both semi-characteristics, you can craft a new TM $O$ which runs them "in parallel, step by step" on the same input $w$. Eventually, one (and only one) of them will halt, and $O$ can accept or reject "w" consequently.

A completely computational-model-agnostic way to state "computing step by step" could be to exploit the Kleene's normal form theorem, where a number $x$ plays an equivalent role to the number of steps. While this is arguably the "true mathematical" way to do it, in computer science we often prefer to work with "programs" (TMs,...) since we find those more familiar.