Controlling the Size of an Open Cover of a Set of Measure Zero
Solution 1:
If the property holds, then $A$ has Hausdorff dimension $0$, because $$\sum_{n=1}^\infty \left(\frac{\varepsilon}{2^n}\right)^d=\frac{\varepsilon^d}{2^d-1}$$ can be made arbitrarily small for each fixed $d>0$ by choosing $\varepsilon$ small enough. The Cantor set has Lebesgue measure $0$ and Hausdorff dimension $\log_3(2)$, so it is a counterexample.