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
As Clayton observed, the space pictured in your question is not the space $X$ defined in William Elliot's original answer.
Let us redefine $X$ in a slightly modified form. For $r \in \mathbb{R}$ let $L_r$ denote the line segment from $(0,0)$ to $(1,r)$ if $r \ge 0$ and the line segment from $(1,0)$ to $(0,r)$ if $r \le 0$. Now define $$X = \bigcup_{q \in \mathbb{Q}} L_q \subset [0,1] \times \mathbb{R} .$$ Note that the original definition was $X = \bigcup_{q \in [-1,1] \cap \mathbb{Q}} L_q$.
All $L_q$ are path connected subspaces of $X$. Hence those with $q \ge 0$ resp. $q \le 0$ belong to the path component of $(0,0)$ resp. $(0,1)$. But $(0,0), (0,1) \in L_0$ which shows that $X$ is path connected.
Let us check that $X$ is not locally connected at any point $x = (x_1,x_2)\in X$. Define $$V(x) = \begin{cases} X \backslash \{(0,1)\} & x_1 = 0 \\ X \backslash \{(0,0)\} & x_1 = 1 \\ X \backslash \{(0,0), (0,1)\} & 0 < x_1 < 1 \end{cases} $$ This is an open neighborhood of $x$ in $X$. For each $r \in \mathbb{R}$ the set $([0,1] \times \mathbb{R}) \backslash L_r$ splits into two disjoint nonempty open sets $O_r^\pm$ above and below $L_r$. Hence the sets $V_r^\pm(x) = V(x) \cap O_r^\pm$ are disjoint nonempty open subsets of $V(x)$. Call $r$ admissible for $x$ if $r$ is irrational and $r < 0$ for $x_1 = 0$ resp. $r > 0$ for $x_1 = 1$. For an admissible $r$ we have $L_r \cap V(x) = \emptyset$ and therefore $V(x) = V_r^+(x) \cup V_r^-(x)$.
Now consider any open $U \subset X$ such that $x \in U \subset V(x)$. We shall show that there exists a regular $r$ such that both $U_r^\pm = U \cap V_r^\pm(x)$ are nonempty. This proves that $U$ is not connected.
Since $U$ is open, the set $S(U) = \{ q \in \mathbb{Q} \mid L_q \cap U \ne \emptyset \}$ is easily seen to be open in $\mathbb{Q}$.
a) $x_1 = 0$. Then $x = (0,q)$ with $q \le 0$. Obviously $q\in S(U)$. Choose $p \in S(U)$ such that $p < q$ and an irrational $r$ such that $p < r < q$. Then $r$ is admissible for $x$ and both $U_r^\pm$ are nonempty.
b) $x_1 = 1$. Can be treated similarly.
c) $0 < x_1 <1$. Choose $p, q \in S(U)$ such that $p < q$ and an irrational $r$ such that $p < r < q$. Then $r$ is admissible for $x$ and both $U_r^\pm$ are nonempty.