Connectedness of $[0,1) \times [0,1]$. I'm taking an introductory course  in topology and I came across the following problem.

Decide if $X=[0,1) \times [0,1]$ with the topology from the lexicographical order is connected or not.

When I started thinking about the problem I didn't have much intuition about this topological space so I tried to prove that see if it was disconnected. For that I had seen before that the space $[0,1) \times \mathbb N$ was disconnected by the open sets $U= \{0\} \times \mathbb N$ and $V=(0,1) \times \mathbb N$. I tried to replicate that same idea. I realized that the set $W \subset X$ given by $\{0\} \times [0,1]$ wasn't an open set so that I couldn't make the same separation as before. This gave me the idea that it was connected but to prove that something is connected it tends to be a little more complicated. I tried to find a suitable homeomorphic topological space but couldn't quite get an idea of which one. I would gladly accept any hints.
 A: $X = [0,1) \times [0,1]$ obeys the conditions that Munkres' states before theorem 24.1 of being a "linear continuum". One can check this essentially the same way as Munkres does (in Example 1 following 24.3) for the whole $[0,1]\times[0,1]$ (our $X$ is the full square minus the final interval $\{1\} \times [0,1]$, which leaves it connected). So $X$ is connected, as it is an order-convex subset of the linear continuum $[0,1]\times [0,1]$. (This is an alternate proof besides checking the linear continuum properties directly. Note that $X$ equals the set of all $(x,y)$ that are $< (1,0)$, e.g.)
$Y=[0,1] \times [0,1)$ is not connected because we're removing a lot of loose interior points, as it were (all points of the form $(x,1)$). $\{0\} \times [0,1)$ is a closed-and-open subset of $Y$, e.g. (all points below $(0,1)$ and above $(0,1)$ from the full square). 
Be careful, the lexicographical order needs careful consideration...
A: Alternatively, you may check that projection to the first coordinate gives a quotient map $q:[0,1)\times [0,1]\to [0,1)$ with connected fibers.
I'm not entirely sure now that $q$ is in fact a quotient map.
Ok, I now think that $q$ is in fact a closed quotient map, but the argument proving it is a bit more involved than what I originally had in mind.
Let $F\subseteq [0,1)\times [0,1]$ be closed and let $(x_\alpha)$ be a convergent net in $q(F)$ with limit $x$; it suffices to see that there exists a point of the form $(x,y)\in F$. If this were not true, then $(x,1)\in F^c$, which is an open set, and any basic open set containing $(x,1)$ must also contain all points of the form $(x+\varepsilon, y)$ for small enough $\varepsilon$. This, and a similar argument around the fact that $(x,0)\in F^c$, proves that $U\times [0,1]\subseteq F^c$ for some standard open neighborhood $U$ of $x$, contradicting the fact that $x_\alpha\to x$. This shows that $q(F)$ is closed.
A: If $<$ is a linear order on a set $X$ and if $Y$ is a non-empty subset of X with an upper bound but no $lub$ then the $<$-order topology on $X$ is disconnected because the set $U_Y$ of upper bounds for $Y$, and the set $X\setminus U_Y,$ are each open and non-empty.
In the Q let $Y=\{0\}\times [0,1).$ Then any $(a,b)\in X$ belongs to $U_Y$ iff $a>0$.  If  $(a,b)\in U_Y$ then $(a,b)>(a/2,0)\in U_Y ,$ so $(a,b)\ne \min U_Y.$ That is, $Y$ has no $lub.$ 
