Let $\mathbb{R}_c$ be $\mathbb{R}$ with the cocountable topology. We'll show that its cone is not locally path-connected, where the cone is $$X=(\mathbb{R}_c \times [0,1])/\sim$$ with $(x,t)\sim (x',t')$ if $t=t'=1$.
Lemma. Let $A$ be an uncountable set with the cocountable topology. Then compact subsets of $A$ are finite and connected subsets are either singletons or uncountable.
Proof. Suppose $B\subset A$ is compact and (countably or uncountably) infinite. Given any countably infinite subset $B_0\subset B$, we can cover $A$ with sets of the form $U_b=b \cup (A \setminus B_0)$ for $b \in B_0$. A finite subcollection of these sets $U_b$ can contain only finitely many elements of $B$, hence $B$ is not compact. Now suppose that $C \subset A$ is connected and contains more than one element. Since finite/countable subspaces in the cocountable topology have the discrete (subspace) topology, $C$ is uncountable. $\square$
Claim 1. Every path into $\mathbb{R}_c$ is constant. Thus no open subsets of $\mathbb{R}_c$ are path-connected; in particular, $\mathbb{R}_c$ is not locally path-connected.
Proof. Let $f:[0,1] \to \mathbb{R}_c$ be continuous. The image $f([0,1])$ is compact and connected, so it must be a singleton set by the lemma. $\square$
Claim 2. The cone on $\mathbb{R}_c$, denoted $X$, is not locally path-connected.
Proof. Fix a point $(x,t)$ in the open subset $\mathbb{R}_c \times [0,1) \subset X$. Any path $f:[0,1] \to\mathbb{R}_c \times [0,1)$ projects to a path $p \circ f:[0,1] \to \mathbb{R}_c$ under the projection $p: \mathbb{R}_c \times [0,1) \to \mathbb{R}_c$. By Claim 1, $p \circ f$ is constant, so all paths into $\mathbb{R}_c \times [0,1)$ have a fixed first coordinate. Because every open neighborhood of $(x,t)$ includes points $(x',t')$ with $x'\neq x$, it follows that no neighborhood of $(x,t)$ is path-connected. Thus $X$ is not locally path-connected. $\square$
Remark. Any uncountable set with the cocountable topology is connected because any two (nonempty) open sets intersect. It follows that open subsets of such a space are themselves uncountable sets with the cocountable (subspace) topology, hence all open subsets are connected. Then certainly the original space is locally connected. Since a finite product of locally connected spaces is locally connected, $\mathbb{R}_c \times [0,1]$ is locally connected. Quotient maps also preserve local connectedness, so this implies that the cone $X$ is locally connected.
Best Answer
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
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.