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喻 良
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Here is a quite short example to show that you question cannot have a positive answer.

Assume that $V$ is a Sacks extension of constructible universe $L$. Then the set of constructible reals $A=(\mathbb{R})^L$ is nonmeasurable and so has Hausdorff dimension 1. But for any $k$, $\sum_k A\subseteq A$ has inner measure 0 and so has no subinterval.

But to construct a counterexample under $ZFC$, we need more knowledge from logic. The rough idea is to construct a nonmeasurable ideal of hyperarithmetic degrees.

喻 良
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