Let $\varphi: [a, b] \longrightarrow \mathbb{R}$ a positive function. There exist $\alpha > 0$ such that the set $X = \lbrace x \in [a,b] | \varphi(x) \geq \alpha \rbrace$ isn't a null set.
$\textbf{My intuitive ideia:}$ Suppose that $\forall \alpha >0$, $X$ is a null set. Since $\varphi$ is positive, $[a, b] = \bigcup X$, therefore, $[a, b]$ is a null set. Contradiction.
It seems informal to me. But I couldn't write formally.
Your idea is almost complete. You just have to make sure you are taking a countable union. A proof could be:
Suppose that for all $\alpha>0$, $X_{\alpha}$ is a null-set. Since $\phi$ is positive, we have $$[a,b]=\bigcup_{n \in \mathbb{N}} X_{1/n}.$$ Since $X_{1/n}$ has measure zero for all $n$ by assumption, we have that $\mu([a,b])=0$, an absurd. $\blacksquare$
Curiously, you do not need measurability of $\phi$ to prove this result.