Why is there no "weakly version" of locally path-connectedness? As usual in topology, there are many different definition for terms like "locally connectedness":
Let $(X,\mathcal{T})$ be  a topological space. Note that I use the following definition of neighbourhood: A set $U\subset X$ is a neighbourhood of $x\in X$, if there is an open set $\mathcal{O}\subset X$, such that $x\in\mathcal{O}\subset X$. Therefore, neighbourhoods can also be closed.
(1) $(X,\mathcal{T})$ is called "weakly locally connected" at $x\in X$, if for every neighbourhood $U\subset X$ of $x$ there is a connected neighbourhood $V$ of $x$, such that $x\in V\subset U$. In othere words $x$ admits a neighbourhood basis of connected sets. If $(X,\mathcal{T})$ is weakly locally connected at every $x\in X$, then it is called "weakly locally connected".
(2) $(X,\mathcal{T})$ is called "locally connected" at $x\in X$, if for every open neighbourhood $U\subset X$ of $x$ there is an open connected neighbourhood $V$ of $x$, such that $x\in V\subset U$. In othere words $x$ admits an open neighbourhood basis of connected sets. If $(X,\mathcal{T})$ is locally connected at every $x\in X$, then it is called "locally connected".
The two definition differ just by the word open.
Obviously, if $(X,\mathcal{T})$ is locally connected at $x\in X$, it is also weakly locally connected at $x$. The reverse is in general not true. However, we can show that every weakly locally connected space is also locally connected. Therefore, the two definition are globally equivalent. (For a proof see https://proofwiki.org/wiki/Equivalence_of_Definitions_of_Locally_Connected_Space)
I am wondering, why there is no such thing as a weakly locally path-connected space. I have found both definition of locally path-connectedness in textbooks: every neighbourhood contains a path-connected neighbourhood and the second one with open neighbourhoods....are they in this case already equivalent locally? Also Wikipedia doesn't make a distinction: They define weakly locally connected and locally connected, but they define only locally path-connected without a weakly version... (https://en.wikipedia.org/wiki/Locally_connected_space)
In other words: Is the following statement true:

Let $x\in X$ be fixed. 
Every neighbourhood $U$ of x has a path-connected neighbourhood $V$,
  such that $x\in V\subset U$
$$\Rightarrow$$
        Every open neighbourhood $U$ of x has an open path-connected neighbourhood $V$, such that $x\in V\subset U$

That the statement is again globally (if we assume that both sides holds for all x) true is obvious, by a similar proof as for locally conectedness..... But I guess that is also true at a certain point, because otherwise there would also ba such a thing as weakly locally path-connectedness.... 
 A: The statement is not true. The space described in this post Consider the "infinite broom" is used to show that weakly locally connected at a point $x$ does not imply locally connected at $x$, but it works equally well to show that weakly locally path-connected at $x$ does not imply locally path-connected at $x$.
The space $X$ can be described as the subset of $\mathbb{R}^2$ consisting of the union $\bigcup_{n,k\in\mathbb{N}} L_{n,k}$ along with the line segment joining $(0,0)$ and $(1,0)$, where each $L_{n,k}$ is the line segment joining the points $(\frac{1}{n+1},\frac{1}{n+1+k})$ and $(\frac{1}{n},0)$. Our distinguished point $x$ will be $(0,0)$. For any neighborhood $U$ of $x$, we can find some $N\in\mathbb{N}$ such that $Y_n=\bigcup_{n\geq N,k\in\mathbb{N}} L_{n,k}\cup [0,\frac{1}{N-1}]\times\{0\}$ is contained in $U$. Then $Y_n$ is a path-connected neighborhood of $x$ (though not open) and thus $X$ is weakly locally path connected at $x$.
However, we cannot find a path-connected open neighborhood of $x$ contained in $U$. Suppose $U$ is a neighborhood of $x$ which does not contain the point $(1,0)$ and that $V$ is a path-connected open neighborhood of $x$. Since $V$ is neighborhood of $x$, it must contain $(\frac{1}{n+1},0)$ for some $n$. Since $V$ is open, it must also contain $(\frac{1}{n+1},\frac{1}{n+1+k})$ for some $k$. And since $V$ is path-connected and contains $(\frac{1}{n+1},\frac{1}{n+1+k})$, it must also contain $(\frac{1}{n},0)$. We may continue in this fashion and conclude that $V$ contains $(1,0)$ and is therefore not a subset of $U$. We did this by only assuming that $V$ was open, a neighborhood of $x$, and path-connected. Thus, $X$ is not locally path connected at $x$.
