When does open and connected imply path-connected?

Solution 1:

(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).

Solution 2:

I think that a topological space X is path connected if and only if the following two properties are true: (a) X is connected (b) every point in X has a path connected neighborhood.

On the other hand, X is locally path connected if and only if any open subset of X has property (b).