Is $f(\sqrt{x}) = \sqrt{f(x)}$ true, where $f(xy) = f(x) + f(y)$ for all positives $x, y \in \mathbb{R}$?
It's obviously false. But the point is that "can it be proved without using the fact that the function satisfies $f(xy) = f(x) + f(y)$ is logarithms?"
All you need to know is that a logarithm is one of the functions that satisfies this equation. In particular, take $f(x) = \log_{10}(x)$ (any choice of logarithm works here). We know that $f(xy) = f(x) + f(y)$, which was the desired property.
However, we note that $$ f(\sqrt{10}) = \frac 12 \neq \sqrt{f(10)} = 1 $$ So, this $f$ is a counterexample to your claim.
Note: if there were no $f$ satisfying the equality except for $f(x) = 0$, then the claim would be true. So, we need to have some kind of example function. If logarithms are the only example functions available, then we need to use a logarithm.