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As stated, countable sets have Hausdorff dimension 0. So any set $S$ with $\mathrm{HD}(S)>0$ has power $\ge \aleph_1$. No need for continuum hypothesis.

Without CH, though, we cannot say whether power $ \ge c = 2^{\aleph_0}$ is required. But this is not about Hausdorff dimension, it is the same question for positive Lebesgue measure in the line. It is consistent with ZFC (follows from Martin's Axiom) that any set with power $< c$ has Lebesgue measure zero, or (for the same reason, or with the same proof, or even consequently) any set with power $< c$ has Hausdorff measure dimension zero. However, without CH (and without Martin's Axiom) there could be sets of reals of power $< c$ but with positive outer Lebesgue measure, and thus Hausdorff dimension 1.

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As stated, countable sets has have Hausdorff dimension 0. So any set $S$ with $\mathrm{HD}(S)>0$ has power $\ge \aleph_1$. No need for continuum hypothesis.

Without CH, thoughtthough, we cannot say whether power $ \ge c = 2^{\aleph_0}$ is required. But this is not about Hausdorff dimension, it is the same question for positive Lebesgue measure in the line. It is consistent with ZFC (follows from Martin's Axiom) that any set with power $< c$ has Lebesgue measure zero, or (for the same reason, or with the same proof, or even consequently) any set with power $< c$ has Hausdorff measure zero. However, without CH (and without Martin's Axiom) there could be sets of reals of power $< c$ but with positive outer Lebesgue measure, and thus Hausdorff dimension 1.

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As stated, countable sets has Hausdorff dimension 0. So any set $S$ with $\mathrm{HD}(S)>0$ has power $\ge \aleph_1$. No need for continuum hypothesis.

Without CH, thought, we cannot say whether power $ \ge c = 2^{\aleph_0}$ is required.