Let f : X → R be a continuous function on a compact metric space...

The compactness, metrizability and other conditions are unnecessary. All that is needed is that f is measurable and $\mu \ll \lambda$, where I use $\lambda$ for Lebesgue measure.
$\lambda \{x\in X:f(x)\ge 0\}=0\wedge \mu \ll \lambda ⟹\mu \{x\in X:f(x)\ge 0\}=0$
And immediately we have that, as X is not $\mu$-null:
${\int }_{X}f\mathrm{d}\mu <0$
I'm afraid we do not quite have:
${\int }_{X}f\mathrm{d}\mu \ge 0⟹f\ge 0\mu \text{ - a.e.}$
Since a function that is both positive and negative can simply cancel, e.g.
${\int }_{\mathbb{R}}\frac{\sin x}{x}\mathrm{d}x$
Famously evaluates to $\pi >0$, but of course $\frac{\sin (x)}{x}$ is negative on a set of infinite measure.
Otherwise, what you said is fine.