Subspace of Lindelöf space is not Lindelöf: Example The Munkres' topology book provides Example 30.5 (p.193, 2nd Ed) for a subspace of a Lindelöf space that need not be Lindelöf as follows:

The ordered square $I_0^2$ is compact; therefore it is Lindelöf
  trivially. However, the subspace $A = I \times (0,1)$ is not Lindelöf.
  For $A$ is the union of the disjoint sets $U_x = \{x\} \times (0,1)$,
  each of which is open in $A$. This collection of sets is uncountable,
  and no proper subcollection covers $A$.

I could largely understand this argument. But it is not clear to me how $U_x$ is open in $A$. Could someone please explain? Thanks.
RD
 A: The set $U_x$ is open, since the topology on the ordered square is not the standard Euclidean topology.  The definition of the order on the square is lexicographic, ie $(x_1,y_1) < (x_2, y_2)$ if either $x_1 < x_2$ or if $x_1 = x_2$ and $y_1 < y_2$.  The corresponding order topology is generated by sets of the form 
$$
\{ (x, y) : (x_1, y_1) < (x,y) < (x_2, y_2)\}
$$
In this example $U_x$ is obtained by taking $(x_1, y_1) = (x,0)$ and $(x_2, y_2) = (x,1)$.
A: A simpler example is, in Munkres' notation, $\overline{S_{\Omega}}$. (The closure of the first uncountable well ordered set in the order topology.)
The space is compact and therefore Lindelöf trivially, (It is compact because it is a well ordered set with a maximum in the order topology, but you can find other ways to prove it's Lindelöf.) but $S_{\Omega}$, the same space without the maximum, is not Lindelöf as the open cover $\{[0,\alpha)\}_{\alpha\in S_{\Omega}}$, has no countable subcover. Note every countable set in $S_{\Omega}$ is bounded.
A: I support Rolf Hoyer's answer. Since I'm reading the same book with you, a quick answer to
"why $U_x=\{\}\times(0,1)$ is open in $= \times(0,1)$ (as a subspace of ordered square)?"
is $\S 16 $ (see Figure 16.1) of the book where shows a counterexample.
When inventing the case of  $= \times(0,1)$, Munkres intended to disconnect all vertical lines (thus they are uncountable) by deleting the lowest/highest horizontal line.
Given

*

*dictionary order $(a\times b, c\times d) $ for $a<c$ and for $b<d$ if $a=c$.


*the space $I \times (0,1)$
We can see that the set $\{ x \}\times(0,1)$ is the "vertical" interval $(x\times 0, x\times1)$ (a form of dictionary order), so the set  $\{\}\times(0,1)$ is an open subset of $ \times(0,1)$ (as a subspace of ordered square) .
