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.
Proof of connectedness
Let $U,V\subseteq \Bbb R^2$ be arbitrary open sets with $L\subseteq U\cup V$ and $L\cap U\cap V=\emptyset$. We want to show that $L\subseteq U$ or $L\subseteq V$.
One of them, wlog. $U$, contains the point $(1,0)$, hence also an open disk $B_r(1,0)$ of some positive radius $r$ around that point. For $N>\frac1r$, we have $(1,\frac1N)\in U$, hence $U$ intersects $L_N$. As $L_N$ is connected (it is homeomorphic with $[0,1]$), we conclude that $L_n\subseteq U$. In particular $(0,0)\in U$. But the $U$ intersects $L_n$ for all $n$, hence $L_n\subseteq U$ for all $n$, i.e., $\bigcup_{n=1}^\infty L_n\subseteq U$. As also $(1,0)\in U$, we have $L\subseteq U$.
Disproof of path connectedness
Assume $L$ were path-connected. Then there exists a path from $(0,0)$ to $(1,0)$, i.e., a continuos map $\gamma\colon [0,1]\to L$ with $\gamma(0)=(0,0)$ and $\gamma(1)=(1,0)$.
Let $t_0=\sup\{\,t\in[0,1]\mid\gamma(t)=(0,0)\,\}$. By continuity of $\gamma$, also $\gamma(t_0)=(0,0)$.
Let $t_1=\inf\{\,t\in[t_0,1]\mid\gamma(t)=(1,0)\,\}$. Again by continuity of $\gamma$, also $\gamma(t_0)=(1,0)$.
Then $p:=\gamma((t_0+t_1)/2)$ is $\ne(0,0)$ and $\ne (1,0)$, hence on one (and only one) $L_n$. But there is no path from $p$ to $(1,0)$ that does not pass through $(0,0)$ (to see this, note that $L\setminus\{(0,0)\}$ is not even connected).
Best Answer
Others have mentioned the topologist's sine curve, that's the canonical example. But I like this one, though it is the same idea:
This is the image of the parametric curve $\gamma(t)=\langle (1+1/t)\cos t,(1+1/t)\sin t\rangle,$ along with the unit circle. There is no path joining a point on the curve with a point on the circle. Yet, the space is the closure of the connected set $\gamma((1,\infty))$, so it is connected.