The part of the question about the continuum hypothesis (CH) seems confused: without assuming (CH) (but assuming axiom of choice so that cardinals work as they should), $\aleph_1$ is by definition the least uncountable cardinal. (The continuum hypothesis asserts that $\aleph_1 = 2^{\aleph_0}$.)

Let $X$ be a metric space, and let $x \in X$. Then it follows immediately from the definition -- see e.g.

http://en.wikipedia.org/wiki/Hausdorff_measure

that for any $d > 0$, the $d$-dimensional Hausdorff measure $H_d(\{x\})$ is equal to zero. (This is just because a point can be covered by a single ball with arbitrarily small diameter.) Since $H_d$ is a measure, it is countably additive: thus $H_d(S) = 0$ for any countable set $S$. If $H_d(S) = 0$, then the Hausdorff dimension of $S$ is at most $d$, so $H_d(S) = 0$ for all
positive $d$ implies that the Hausdorff dimension of $d = 0$.

Possibly you wanted to ask: without assuming CH, does a set of positive Hausdorff dimension necessarily have at least continuum cardinality? (I don't know the answer.)