For every $B \subset \mathbb{R}^n$ one has $\dim_H B \le n$, where $\dim_H$ is the Hausdorff dimension. If B contains an open set or a set of strictly positive Lebesgue measure, then $\dim_H B=n$.
I am looking at the following solution to this problem from Rene Schilling.
Solution. We know that $0 \le \dim_H B \le n$. If $B$ contains an open set $U$ (or a set of non-zero Lebesgue measure), we see $H^n(B) \ge H^n(U) > 0$; intersect with a large open ball $K$ to make sure that $H^n(B \cap K) < \infty$ and $U \cap K \subset B \cap K$. THis shows $n = \dim_H(B\cap K)\le \dim_H (B) \le n$.
However, I don't see why we need to consider a large open ball $K$ in this case. We already have $0<H^n(B)<\infty$, so by the definition of the Hausdorff dimension, i.e. $$\dim_H(B) = \sup\{s \in (0,\infty): H^s(B)>0\}$$ don't we immediately have that the dimension is $n$?
But it is not given that $H^{n}(B)<\infty$, it could be $H^{n}(B)=\infty$, so we need to intersect with a larger ball $K$ to make sure that $H^{n}(B\cap K)<\infty$, and we also have $H^{n}(B\cap K)>0$, then $\dim_{H}H^{n}(B\cap K)=n$.