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.
The fact that $X$ is locally connected iff for all open subsets $O$ of $X$, all components of $O$ (as a subspace) are open in $O$ (and $X$ too), should be known to you. It's Munkres theorem 25.3
So we will show $X$ to be locally connected using the right to left direction of this theorem.: let $O$ be open in $X$ and let $C$ be a component of $O$ (in the subspace topology).
Let $x \in C$. Then $x \in O$ and $O$ is a neighbourhood of $x$ so by assumption of weak locally connectedness there is a connected set $C_x \subseteq O$ such that $x \in \operatorname{int}(C_x)$ (as $C_x$ is a a neighbourhood of $x$).
Then $C$ and $C_x$ are both connected subsets of $O$ that intersect (in $x$) so
$C \cup C_x$ is also a connected subset of $O$. As $C$ is a component of $O$ and as such is a maximally connected subset of $O$,
$$C \cup C_x = C \text{ so } x \in \operatorname{int}(C_x) \subseteq C$$
showing finally that $x$ is an interior point of $C$. As this holds for all $x \in C$, $C$ is open, as required.
Best Answer
"Weakly locally path-connected" is actually equivalent to "locally path-connected". To prove this, suppose $X$ is weakly locally path-connected, $x\in X$, and $U$ is an open neighborhood of $x$. Let $V$ be the path-component of $x$ in $U$. I claim that $V$ is in fact open, and so is a path-connected open neighborhood of $x$ contained in $U$. To prove this, suppose $y\in V$. By weak local path-connectedness, there exists an open set $W$ such that $y\in W\subseteq U$ and every element of $W$ is connected to $y$ by a path in $U$. Then every element of $W$ is in the path-component $V$. So $V$ contains a neighborhood of each of its points, and is open.