Exercise: Let $L_n:=\{(x,\frac{x}{n})\mid0\leq x\leq1\}\subset\mathbb{R}^2$ for $n\in\mathbb{Z}_+$. Then consider $X=\{(1,0)\}\cup\bigcup_{n=1}^\infty L_n\subset\mathbb{R}^2$ together with the subspace topology. Is this topological space path-connected, connected, and/or locally connected? Explain.
Thoughts: I'm a bit unsure on how to proceed with the proof of the path-connectedness. If we show that any continuous map $f:[0,1]\rightarrow X$ with $f(0)=p$ is constant by showing that $f^{-1}(\{p\})$ is clopen, then I think that it would follow, but I'm unsure of exactly how. We probably also need to use that the continuous image of a connected set is connected.
I don't think that it's locally connected as the lines given by $L_n$ don't get arbitrarily close to each other, which means that we can find a local separation regardless of how close we come to any point along those lines.
HINT: I’m assuming that your $p$ is the point $\langle 1,0\rangle$. Note that $X\setminus\{p\}$ is locally connected and path connected, so a good first step is to show that $p$ has no local base of connected open nbhds. Is there are connected open nbhd of $p$ contained in $U\cap X$, where $U$ is the open ball of radius $\frac12$ centred at $p$?
To show that $X$ isn’t path connected, suppose that $f$ is a path from $p$ to the origin, so that $f(0)=p$. $f^{-1}[\{p\}]$ is a compact subset of $[0,1]$, so it has a maximum element, and by a minor modification of $f$ we may assume that $f^{-1}[\{p\}]=\{0\}$. (Why?)
Let
$$V=\left\{\langle x,y\rangle\in X:x>\frac12\right\}\,;$$
$V$ is an open nbhd of $p$, and $f$ is continuous, so there is an $a\in(0,1]$ such that if $U=[0,a)$, then $f[U]\subseteq V$. $U$ is an interval, so $f[U]$ is a connected subset of $X$ containing $p$. Let $\left\langle x_0,\frac{x_0}{n_0}\right\rangle\in f[U]\setminus\{p\}$, and show that
$$W=\left\{\left\langle x,\frac{x}{n_0}\right\rangle\in f[U]:x>\frac12\right\}$$
and $f[U]\setminus W$ form a separation of $f[U]$, contradicting the connectedness of $f[U]$.