is the lexicographic order topology on the unit square connected/path connected? I was wondering, given the lexicographic order topology on $S=[0,1] \times [0,1]$, is it connected (and path connected)? 
I found a reference to Steen's and Seebach's Counterexamples in Topology, and in page 73 they say that: 

Since in the linear order on $S$ there are no consecutive points, and since every (bounded) subset of $S$ has a least upper bound, $S$ is connected.

But I don't know what a consecutive point is? (perhaps there is another name for this type of point)
 And I dont see how this implies that $S$ is connected? 
And my second question is - is $S$ path connected? According to the book it isn't, but I don't see exactly how.

Suppose $X$ ($=S$?) is path connected, so there exists a continuous path $\gamma : [a,b] \to X$ such that $\gamma(a)=(0,0)=P$ and $\gamma(b)=(1,1)=Q$. Since every point of $X - \{P,Q\}$ disconnects $P$ from $Q$ it follows that $\gamma$ is surjective. For each $t \in [0,1]$ let $J_t$ be the open vertical segment with lower endpoint at $(t,1/4)$ and with upper endpoint at $(t,3/4)$. The sets $J_t$ are open, nonempty, and pairwise disjoint in $X$. It follows that the sets $\gamma^{-1}(J_t)$ are open, nonempty, and pairwise disjoint in $[a,b]$. But this is impossible, because $[a,b]$ contains a countable dense set.
Almost the same proof shows that the path components of $X$ are precisely the vertical arcs $\{t\} \times [0,1]$.
