show that the ordered square is locally connected~

The argument given at Onesidey (your link) is about as straightforward as you can get; I don’t offhand know of a different one, let alone a simpler one.

Let $U$ be an open nbhd of the point $p=\left\langle\frac12,0\right\rangle$. Because the lexicographically ordered square has the order topology, $U$ must contain an open interval around $p$ in the lexicographic order.

Let $\langle x_n:n\in\Bbb N\rangle$ be a strictly increasing sequence of positive real numbers converging to $\frac12$, and let $\langle y_n:n\in\Bbb N\rangle$ be a strictly decreasing sequence of real numbers less than $1$ converging to $0$. For $n\in\Bbb N$ let $a_n=\langle x_n.0\rangle$ and $b_n=\left\langle\frac12,y_n\right\rangle$. Then the sequences $\langle a_n:n\in\Bbb N\rangle$ and $\langle b_n:n\in\Bbb N\rangle$ converge to $p$ in the lexicographic order topology, and

$$a_0\prec a_1\prec a_2\prec\ldots\prec p\prec\ldots\prec b_2\prec b_1\prec b_0\;.$$

Thus, there is some $n\in\Bbb N$ such that the open interval $(a_n,b_n)$ in the lexicographic order topology is a subset of $U$: if $a_n\prec u\prec b_n$, then $u\in U$. This open interval contains all points $\langle x,y\rangle$ in the square such that

• $x_n<x<\frac12$, or
• $x=x_n$ and $y>0$, or
• $x=\frac12$ and $0\le y<y_n$.

Here’s a rough sketch:

The black point is $p$ (and is so labelled); the blue points are $p_n=\langle x_n,0\rangle$ to the left of $p$ and $b_n=\left\langle\frac12,y_n\right\rangle$ above $p$; and the open interval $(a_n,b_n)$ is colored orange. It includes all of the points in the square strictly above the blue point $a_n$, all of the points of the square between the lines $x=x_n$ and $x=\frac12$, and all of the points in the square strictly below the blue point $b_n$.

Every open nbhd of $p$ contains an open interval of this kind, because every open interval around $p$ in the lexicographic order contains one of this kind.

Now pick real numbers $s$ and $t$ such that $x_n<s<t<\frac12$, and let

$$R=\{\langle x,y\rangle:s\le x\le t\text{ and }0\le y\le 1\}\;;$$

this set is the closed black rectangle in the picture below.

In terms of the lexicographic order, $R$ is the closed interval from $\langle s,0\rangle$ to $\langle t,1 \rangle$, and it’s homeomorphic to the whole square, which is the closed interval from $\langle 0,0\rangle$ to $\langle 1,1\rangle$.

In fact, let

$$h:[s,t]\to[0,1]:x\mapsto\frac{x-s}{t-s}\;;$$

you can easily check that $h$ is a homeomorphism from the ordinary closed interval $[s,t]$ in $\Bbb R$ to the closed interval $[0,1]$, and it’s not much harder to verify that if $X$ is the lexicographically ordered square, then the map

$$\varphi:R\to X:\langle x,y\rangle\mapsto\langle h(x),y\rangle$$

is a homeomorphism.

This shows that every nbhd of $p$ does indeed contain a set homeomorphic to $X$, and since $X$ is not path connected (as shown by Munkres in his Example $24.6$), no neighbourhood of $p$ can be path connected, and $X$ is therefore not locally path connected.

Instead of this particular point $p$ we could actually have used any point $\langle x,0\rangle$ with $0<x\le 1$ or any point $\langle x,1\rangle$ with $0\le x<1$, i.e., any point along the top or bottom edge of the square except the endpoints of the lexicographic order.