It seems that the following statement is pretty simple to prove:
Let $\emptyset \neq M$ be an arbitrary open set in $\mathbb{R}^n$, where n is the smallest integer such that $M \subseteq \mathbb{R}^n$. Then $dim_H(M) = n$, for $M$ contains an n-dimensional ball.
Could you give me hints in order to do so?
Source: Massopoust, Interpolation and Approximation with Splines and Fractals.
Hausdorff dimension is monotonic as Why is Hausdorff dimension monotonic? explains. If we show that $dim_H \; \mathbb{R}^n = dim_H \; B(x,r) = n$ then for any open set we have $B(x,r) \subseteq O \subseteq \mathbb{R}^n$. And thus $dim_H \; O = n$.
For the computation of the Hausdorff dimension of a ball, I sketch the following argument. In the plane one may see that the optimal way of filling a squares with squares of side $\epsilon$ requires $\frac{1}{\epsilon^2}$ squares. Generalizing to dimension $n$ we need $\frac{1}{\epsilon^n}$ squares. If we compute $H^s_{\epsilon}$ of the square we then get $\epsilon^{s-n}$. Taking the limit when $\epsilon \to 0$ it is clear that the critical value defining the $0 - \infty$ behaviour is $s = n$.
The computation of the Hausdorff dimension of $\mathbb{R}^n$ is described in Hausdorff dimension of $\mathrm{R}^d$.