The most likely reason is that it is less clear what happens in neighborhoods of $(0,0)$ compared to what happens in neighborhoods of $(0,y)$ for $y\neq 0$. The author is only trying to argue that the space as a whole is not locally connected so does not care whether or not the space is locally connected at $(0,0)$. The author likely felt (either correctly or incorrectly) that more of an argument would be required to demonstrate the lack of local connectivity at this point.
But, you can rest assured knowing that the space is in fact not locally connected at $(0,0)$ and the same argument can be applied here.
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 intuition behind the several examples constructed for you here is as follows: take some space that you know to have many disconnected components that are 'very close' to each other. Examples include $\mathbb{Q}$ and $\{0,1/n, n\geq1\}$ (for which the disconnected components are the one-element subsets). These spaces are not locally connected because of the 'closeness' of the components: in the cases above, the space is not locally connected at $x = 0$ because no matter how far we zoom in on zero, we cannot isolate it from the connected components around it.
These spaces are of course not connected either, but we can make them so by joining all the components together somewhere far away. For instance, for the examples above, we could take the point $(0,1)$ in the plane (above the real line) and draw a line from this point to every point in our original space. This new space is connected because we can use these lines to move along a continuous path from any point to any other point. But it remains not locally connected since when we look at the vicinity of the points in the original space, we do not see this 'connection point' that we've added. All of the examples here operate along these lines.