Any open interval in the ordered square is a linear continuum, therefore, it is connected. At the same time any open set that contains $(x,0)$ for $x>0$ contains some point $(y,0)$ , $y<x$ , and the argument similar to Example 6 of §24 shows that it is not path connected. Similar for points $(x,1)$ , $x<1$ . The space is locally path connected at all other points. (Another way to prove that it is not locally path connected is as follows: if it were locally path connected then its components would be the same as its path components, i.e. it would be path connected which contradicts Example 6.) The path components are the vertical segments. By the way, in the subspace topology as a subspace of $\mathbb{R}^2$ in the dictionary topology (which is strictly finer than the order topology of the ordered square, remember exercise 10 of §16?) it is locally path connected but still not path connected (it is not even connected in this topology).