One counterexample is a variant on the famous topologist's sine curve.
Consider the graph of $y = \sin(\pi/x)$ for $0<x<1$, together with a closed arc from the point $(1,0)$ to $(0,0)$:
This space is obviously path-connected, but it is not locally path-connected (or even locally connected) at the point $(0,0)$.
(this is still a partial answer)
The following two properties of a topological space $X$ are equivalent (I use your numeration):
(2) every path-connected component of an open subset of $X$ is open
(3) $X$ is locally path-connected
Proof.
(3) implies (2). Take a point $x$ of an open subset $U$, and consider the path-connected component $Y$ of the subset that contains $x$; since the space is locally path-connected, there exists a path-connected open neighbourhood $V$ of $x$ contained in $U$; obviously $V$ must be contained in $Y$, so $Y$ is open.
(2) implies (3). Take a point $x$ and an open neighbourhood $U$ of it. By hypothesis the path-connected component of $U$ containing $x$ is open so we are done.
The following property is not equivalent to (2)-(3), but is implied by them:
(1) open connected subsets of $X$ are path-connected.
As Nate suggested, the set of the rational number with the euclidean topology trivially satisfies (1) but doesn't satisfy (2)-(3).
Best Answer
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".