The answer is yes under the additional assumption that the set is compact and I do not know what happens in the general case. The result is a consequence of the following one, see [1] and references therein.

Theorem. If a compact set $A\subset\mathbb{R}^n$ has non-$\sigma$-finite measure $\mathcal{H}^\beta$, then there us a subset $B\subset A$ such that $0<\mathcal{H}^\beta<\infty$.

[1] R.O. Davies, A theorem on the existence of non-σ-finite subsets. Mathematika 15 (1968), 60–62.

The answer is yes under the additional assumption that the set is compact and I do not know what happens in the general case. The result is a consequence of the following one, see [1] and references therein.

Theorem. If a compact set $A\subset\mathbb{R}^n$ has non-$\sigma$-finite measure $\mathcal{H}^\beta$, then there exists a subset $B\subset A$ such that $0<\mathcal{H}^\beta(B)<\infty$.

[1] R.O. Davies, A theorem on the existence of non-σ-finite subsets. Mathematika 15 (1968), 60–62.

The answer is yes and the result is a consequence of much more general one. For the following one, see [1] and references therein.

Theorem. If a compact set $A\subset\mathbb{R}^n$ has non-$\sigma$-finite measure $\mathcal{H}^\beta$, then there us a subset $B\subset A$ such that $0<\mathcal{H}^\beta<\infty$.

[1] R.O. Davies, A theorem on the existence of non-σ-finite subsets. Mathematika 15 (1968), 60–62.

The answer is yes under the additional assumption that the set is compact and I do not know what happens in the general case. The result is a consequence of the following one, see [1] and references therein.

Theorem. If a compact set $A\subset\mathbb{R}^n$ has non-$\sigma$-finite measure $\mathcal{H}^\beta$, then there us a subset $B\subset A$ such that $0<\mathcal{H}^\beta<\infty$.

[1] R.O. Davies, A theorem on the existence of non-σ-finite subsets. Mathematika 15 (1968), 60–62.