When does open and connected imply path-connected? It's well known that, in $\mathbb{R}^n$:

(1) Open and Connected $\Rightarrow$ Path-connected

The proof essentially goes through the fact that (2) Every path-connected component will be open. Using this fact, we arrive at a contradiction if we suppose there are more than one path-connected component.
Well, trying to see where (1) would keep being valid, I arrived at the following:

If $X$ is a locally convex topological vector space, then (1) is valid.

But, to prove this, I proved (2). But for that, you use the (rather strong imo) convexity of a local base (to essentially repeat the argument for $\mathbb{R}^n$). But I'm not satisfied with this, as I think that we are using a lot of strong conditions (using the existence of a "line segment" to prove the existence of a "curve" seems very bazooka-like to me).
Anyway, given the previous considerations, my questions are:
Are there more general examples of spaces where (1) holds? Is there a characterization of spaces that satisfy this property?
And bonus question, since (2) implies (1) if the set is open.:
Are there more general examples of spaces where open sets satisfy (2)? Is there a characterization spaces that satisfy this property?
 A: (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).  
A: 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).
