In algebraic geometry, one defines the dimension at a point of a variety $X$ as the length of the longest chain of irreducible closed subsets (in the Zariski topology of the variety) containing the point. This definition is purely topological, yet it is useless for Hausdorff spaces since for them the only irreducible closed subsets are singletons.
There is, however, another feature of the Zariski topology: it is locally connected and locally irreducible, which allows us to rephrase the definition of dimension at a point $x\in X$ as the largest $n$ for which there is a chain of connected, locally connected, closed nowhere dense (i.e. with empty interior) sets $x\in C_1\subsetneq C_2\subsetneq\dots C_n\subseteq X$ with each $C_i$ nowhere dense in $C_{i+1}$ (in the subspace topology). For lack of a standard term that I'm aware of, I would call this the "(connected) nowhere dense dimension of $x\in X$".
My question is whether $\mathbb R^n$ with its standard Euclidean topology has finite "nowhere dense" dimension $n$. Unless I am mistaken, it is certainly the case that $\mathbb R$ itself has "nowhere dense dimension" $1$ (notice the connectedness is crucial). Does $\mathbb R^2$ have "nowhere dense dimension" $2$? It seems to me it might.
No, $\mathbb{R}^2$ has nowhere dense dimension higher than 2. In particular, it's easy to extend the following example with a "fractal" construction to show that it has infinite nowhere dense dimension.
Take $C_0=\{(0,0)\}$, $C_1=[0,1]\times\{0\}$ an horizontal segment, $C_3=[0,1]\times[0,1]$ a square, now we want to construct $C_2$.
For every rational number $p/q\in [0,1]$ such that $p/q$ is the minimal form (i.e. $(p,q)=1$, $p,q\geq 0$) consider the vertical segment $S_{p/q}=\{p/q\}\times [0,1/q]$. Call $C_2$ the union of $C_1$ and $S_{p,q}$ for every rational $p/q\in [0,1]$: it's easy to see that $C_2$ is closed and connected, and that both the inclusions $C_1\subseteq C_2$ and $C_2\subseteq C_3$ are nowhere dense. Moreover, $C_2$ is clearly locally connected at the points of $C_1$.
Hence, take a point $x=(p/q,y)\in S_{p/q}\setminus C_1$, $y\neq 0$. There exists an $r>0$ such that if $0<|p'/q'-p/q|<r$, then $1/q'<y/2$. Now choose $r'=\operatorname{min}\{y/2,r\}$. One can see that $B(x,r')\cap C_2=B(x,r')\cap S_{p/q}$, and $S_{p,q}$ is locally connected, hence $C_2$ is locally connected too.