“locally” operator in “locally path-connected space” 
For any property p of topological spaces, p implies locally p.

topospaces.subwiki.org. Locally operator

Locally path-connected space … This property is obtained by applying
  the locally operator to the property: path-connected space

topospaces.subwiki.org. Locally path-connected space

This space is obviously path-connected, but it is not locally
  path-connected

math.stackexchange.com
This seems like a contradiction.
 A: The definition in the first link: property $p$ holds locally if for each $x \in X$ there exists a neighborhood $U \ni x$ such that $p$ holds on $U$.
Then it is obvious that if $p$ holds (on the full space) it will also hold locally, just take $U = X$. An example of such a property is compactness.
Now the definition of local (path-)connectedness uses a different (stronger) notion of "locally", as Yuki spelled out. In the first link this is called a "strongly locally operator".
A: Locally path-connected means that for every $x\in X$ and every neighbourhood $V$ of $x$, you can find an path-connected neighbourhood $U$ of $x$ such that $U\subset V$ (and not only "for each $x\in X$ there exists an path-connected neighbourhood of $x$").
Defining in this form, it's possible to have an path-connected space which isn't locally path-connected.
A: In this page they explain the difference. For an example of path-connected space which is not strongly locally path-connected, consider the subspace of $[0,1] \times [0,1]$ of points whose $y$ coordinate is rational with the subspace topology. It is path-connected (use a path going to $x$ coordinate $0$ or $1$ to connect two points), but a small enough neighborhood of a point with $x$-coordinate not equal to $0$ or $1$ will not have a path-connected neighborhood because intersecting an open ball (in $[0,1] \times [0,1]$) with this subspace will give you a bunch of small line segments very close to each other, but not path-connected. 
Hope that helps, 
