$\DeclareMathOperator{\diam}{diam}\DeclareMathOperator{\dim}{dim}$
No, this cannot occur.
First, let's quickly recall the definition and set some notation. For $X$ a metric space we have:
A $r$-cover of $X$ is a countable set of open balls in $X$ each of radius $<r$, whose union is all of $X$.
If $\mathcal{C}$ is a countable cover of $X$ by open balls and $a>0$, we let $\diam^a(\mathcal{C})$ be the sum of the $a$th powers of the radii of the balls in $\mathcal{C}$.
For $d>0$, we set $\mathcal{H}^d(X)$ to be the limit, as $r\rightarrow 0$, of the infimum of $\diam^d(\mathcal{C})$ for $\mathcal{C}$ an $r$-cover of $X$.
Finally, $\dim_H(X)=\inf\{d: \mathcal{H}^d(X)=0\}$.
The key point is that if $X$ is a subspace of $Y$ then every cover of $Y$ restricts to a cover of $X$, which in turn gives us $$\mathcal{H}^d(X)\le\mathcal{H}^d(Y)$$ for each $d$. Consequently we have $$\{d: \mathcal{H}^d(X)=0\}\supseteq \{d: \mathcal{H}^d(Y)=0\},$$ and this gives us the crucial result that dimension only increases when we add points: $$\color{red}{X\subseteq Y\implies \dim_H(X)\le \dim_H(Y).}$$
This immediately tells us that since $\mathbb{R}^2$ has Hausdorff dimension $2$, any subset of $\mathbb{R}^2$ has Hausdorff dimension $\le 2$.
First off, the term "fractal" does not have a universally agreed upon meaning. I have written about this before on Math SE, so I won't go into that again. I will point out that Mandelbrot walked back from the definition you provide in the second edition of The Fractal Geometry of Nature, so even he was not entirely wed to that notion.
Inductive Dimensions
Regarding the meat of your question, you cite three different topological notions of dimension: the small inductive dimension, the large inductive dimension, and the Lebesgue covering dimension. For the sake of completeness, let us recall the definitions (note: this presentation assumes familiarity with some of the basics of point-set topology, i.e. the definitions of open and closed sets, limit points, boundaries, etc—these definitions are cribbed from James Robinson's Dimensions, Embeddings, and Attractors):
Definition: The small inductive dimension, denoted $\DeclareMathOperator{ind}{ind}\ind(\cdot)$, is defined as follows:
- $\ind(\varnothing) = -1$,
- $\ind(X) \le n$ if for every point $p \in X$, $p$ has "arbitrarily small" neighborhoods $U$ with $\ind(\partial U) \le n-1$, where $\partial U$ denotes the boundary of $U$, and
- $\ind(X) = n$ if $\ind(X) \le n$ but it is not true that $\ind(X) \le n-1$.
The large inductive dimension, denoted $\DeclareMathOperator{Ind}{Ind}\Ind(\cdot)$ is defined similarly, with (2) replaced by
- $\Ind(X) \le n$ if for every closed set $A \subseteq X$ and each open set $V \subseteq X$ which contains $A$ there exists an open set $U \subseteq X$ such that
$$ A\subseteq \overline{U} \subseteq V \qquad\text{and}\qquad \Ind(\partial U) \le n-1. $$
Both of these definitions are pretty gnarly, but they are trying to get at the same essential idea: we can understand the dimension of a set by looking at the boundaries of subsets; the dimension of a space should be greater than the dimension of the boundary of an open subset of the space.
Thinking in a very Euclidean way, an open set in $\mathbb{R}^2$ is (more or less) a disk. The boundary of a disk is a circle, so $\mathbb{R}^2$ should be one dimension greater than a circle. An open set in the circle is (roughly) an interval, and the boundary of an interval consists of two points. Thus the circle should have dimension one greater than a set containing two points. The boundary of a discrete set is empty, and so a collection of discrete points should have dimension one greater than the emptyset. Thus
\begin{align}
\DeclareMathOperator{Dim}{dim}
\Dim(\mathbb{R}^2)
&= 1 + \Dim(\text{a circle}) \\
&= 1 + (1 + \Dim(\text{two points})) \\
&= 1 + (1 + (1 + \Dim(\varnothing))) \\
&= 1 + (1 + (1 + -1))) \\
&= 2,
\end{align}
where $\Dim$ denotes some appropriate notion of dimension.
The precise definitions of the small and large inductive dimensions try to capture this idea in a more general way, with slight variations in the way in which the open sets of concern are chosen. As is often the case, the jump from intuitive to precise requires a great deal of work.
Covering Dimension
The Lebesgue covering dimension approaches things slightly differently:
Definition: The order of an open covering of a topological space is the largest $n$ such that there exist $n+1$ sets in the covering with non-empty intersection. An open covering $\mathscr{V}$ is a refinement of an open covering $\mathscr{U}$ if every member of $\mathscr{V}$ is contained in some element of $\mathscr{U}$.
A set $A$ has Lebesgue covering dimension less than or equal to $n$, denoted $\Dim_{L}(A) \le n$, if every covering has a refinement of order less than or equal to $n$. $\dim_{L}(A) = n$ if $\dim_{L}(A) \le n$, but not $\Dim_{L}(A) \le n-1$.
Again, there is a pretty intuitive observation about Euclidean space running around behind the scenes. Imagine trying to cover $\mathbb{R}$ by open intervals, and then throwing away as many sets as you can. What you will likely end up with is a collection of intervals which overlap "just a little" at their ends, see the figure.
No matter how gnarly the initial cover is, I can always find a way to "throw away" enough sets so that the remaining sets don't overlap too much. Specifically, no point on the real line will be contained in more than two of the covering intervals.
By contrast, if you attempt to cover the plane $\mathbb{R}^2$ by disks such that no point is contained in more than two disks, you will rapidly find that this is impossible—there must be some points contained in three disks.
Further Reading
Robinson, James C., Dimensions, embeddings, and attractors, Cambridge Tracts in Mathematics 186. Cambridge: Cambridge University Press (ISBN 978-0-521-89805-8/hbk). xii, 205 p. (2011). ZBL1222.37004.
Engelking, Ryszard, Dimension theory. A revised and enlarged translation of ”Teoria wymiaru”, Warszawa 1977, by the author, North-Holland Mathematical Library. Vol. 19. Amsterdam, Oxford, New York: North-Holland Publishing Company. Warszawa: PWN - Polish Scientific Publishers. X, 314p. $ 44.50; Dfl. 100.00 (1978). ZBL0401.54029.
Hurewicz, W.; Wallman, H., Dimension theory., 165 p. Princeton University Press (1941). ZBL67.1092.03.
Best Answer
A "curve" is a function. The "Peano curve" is a function whose domain is $[0,1]$ and whose range is $[0,1]^2$. But "topological dimension" and "Hausdorff dimension" apply only to metric spaces. So when you say "the topological or Hausdorff dimension of $X$", then $X$ must be a metric space.
So the question arises: what is meant by "the topological or Hausdorff dimension of the Peano curve"?