"$\mathbb{R}^2$ can't be totally ordered" (nicely)

I have not seen a standard definition of a nice total ordering, but I like the one you provided.

Neither $\mathbb{R}^2$ nor $S^1$ admit a nice ordering by your definition. The proof is straightforward:

Claim 1: In a topological space $X$ with a nice total ordering $<$, for all $x\in X$ the sets $\{y \mid y<x\}$ and $\{y \mid y> x\}$ are open.

Claim 2: For $X = \mathbb{R}^2$ or $X=S^1$, there is no way to write $X=A\cup B\cup\{x\}$ for disjoint nonempty open sets $A,B$.

The proof of each claim is straightforward: For the first, note that the set $\{y\mid y<x\}$ can be written as a union of open balls around all points inside itself, and similarly for $\{y \mid y>x\}$. For the second, observe that the compliment of a singleton set in $\mathbb{R}^2$ or in $S^1$ is connected.

Suppose you have a point $x_0 \in X$ such that $X \setminus \{ x_0 \}$ is connected. Then under your definition of a "nice" total order, $\{ x \in X \mid x < x_0 \}$ and $\{ x \in X \mid x_0 < x \}$ will be a partition of $X \setminus \{ x_0 \}$ into two disjoint open subsets; therefore, one of them must be the whole subspace $X \setminus \{ x_0 \}$. In other words, $x_0$ must be either a maximum or a minimum element of $X$ under the given order.

Now, if you can find at least three distinct points $x_0, x_1, x_2$ such that $X \setminus \{ x_i \}$ is connected for each $i \in \{ 0, 1, 2 \}$, then you will get a contradiction. This is the case both for $X = \mathbb{R}^2$ and for $X = S^1$.