Hausdorff dimension of a connected set in $\mathbb{R}^d$ I recently asked in this question, whether a set $A$ not being  totally disconnected  in $\mathbb{R}$ implies $\dim(A)=1$? The answer I was given, showed that the answer is yes by relatively simple arguments.
My follow-up question, is

Does connectedness of $A\subseteq \mathbb{R}^d$ imply a lower estimate on the dimension of $\dim (A)$, in $\mathbb{R}^d$ for $d>1$ when $\vert A \vert\geq 1$?

whether connectedness implies a lower estimate on the dimension of a set in $\mathbb{R}^d$ for $d>1$?
I think that if $A$ is path connected, then $\dim(A)\geq 1$, but to conclude this I need to be sure that curves in $\mathbb{R}^d$ satisfy $\dim\geq 1$. Since the arguments in $1$-dimension were simpler than I suspected, I thought that perhaps the arguments here would be relatively simpler as well?
Is there a lower estimate on $\dim(A)$ when $A$ is path connected? Is there also an estimate when $A$ is just connected?
 A: Theorem: if $A \subseteq \mathbb{R}^d$ is connected and $|A| > 1$, then $\operatorname{dim}(A) \geqslant 1$.
Proof. Without loss of generality $0 \in A$ and $p \in A$, where $p = (1, 0, \ldots, 0)$. It suffices to show that for any $d < 1$ we have that $\mathcal{H}^d(A) = \infty$ where
$$\mathcal{H}^d(A) = \lim_{r \to 0^+} \inf \left\{ \sum_{i=1}^{\infty} r_i^d : A \subseteq \bigcup_{i=1}^{\infty} B(x_i, r_i) \text{ for some } x_i \in \mathbb{R}^d, \ 0 < r_i < r \right\}.$$
Fix $r \in (0, 1)$ and consider any family of balls $\left< B(x_i, r_i) : i \in \mathbb{N} \right>$ covering $A$ where $0 < r_i < r$. Denote by $\pi : \mathbb{R}^d \to \mathbb{R}$ the projection onto the first coordinate.
By the connectedness of $A$, for each $t \in (0, 1)$ the hyperplane $\{ t \} \times \mathbb{R}^{d-1}$ must intersect $A$ in some point $p_t$ and this point is covered by some ball $B(x_i, r_i)$. Therefore  the family $\mathcal{I} = \left< \pi \big[ B(x_i, r_i) \big] : i \in \mathbb{N} \right>$ is a covering of $(0, 1)$ with one-dimensional balls of radii $0 < r_i < r$. But we know that $\mathcal{H}^d \big( (0, 1) \big) = \infty$ (because $d < 1$), hence for sufficiently small $r$ we must have that $\sum \limits_{i=1}^{\infty} r_i^d$ is arbitrarily big. That finishes the proof. $\blacksquare$
