Proof of Cauchy's functional equation

I came across Cauchy's functional equation recently and am trying to understand the proofs for it - there are many, but almost all of which are very difficult for me to understand. I know that solutions to the equation $$f(x + y) = f(x) + f(y)$$ are of the form $$f(q) = cq\ \forall\ c \in \mathbb{Q}.$$

Wikipedia provides a very nice proof of it using just elementary algebra here.

I understand all of the proof except for one line, which has got to do with the case where $q > 0$ is considered. I will re-write (part of) the proof below for convenience.

If $q > 0$, then $f(qx) = qf(x)\ \forall\ q \in \mathbb{Q}^+$.

$\implies f(q) = qf(1)$

I know the proof goes from $f(qx) = qf(x)$ to $f(q) = qf(1)$ by letting $x = 1$, but why is this sufficient for a proof? I mean, if I let $x$ take another value, then $c$ would be a different constant right? Doesn't the proof have to be general, in the sense that it covers all values of $x$?

This seems like a trivial observation and maybe it is - it has been a long day of me staring at numbers, but anyone kind enough to explain how it can be concluded that $f(q) = qf(1)$ will be greatly appreciated :)

You just set $x= 1$ in your relation $f(qx) = q\,f(x)$.

Edit: To show some function is linear (over $\mathbb{R}_+$, for example), it suffices to show $f(x) = c\,x$, and of course, $c = f(1)$.