Prob. 2, Sec. 28 in Munkres' TOPOLOGY, 2nd ed: Compactness of $[0,1]$ in the lower limit topology Let $\mathbb{R}_l$ denote the set of real numbers with the topology having as a basis all the half open intervals $[a,b)$ on the real line. 
Then is the closed interval $[0,1]$ compact as a subspace of $\mathbb{R}_l$? 
My effort: 

Let 
  $$A \ \colon= \ \left\{ \ {1 \over 2 } - {1 \over {n+1}} \ \colon \ n \in \mathbb{N} \ \right\} = \left\{ \ 0, \frac{1}{6}, \frac{1}{4}, \frac{3}{10}, \frac{1}{3}, \frac{5}{14}, \frac{3}{8}, \frac{7}{18}, \frac{2}{5}, \ldots \right\}.$$
Then $A$ is an infinite subset of $[0,1]$, but, as we show below, $A$ has no limit points in $[0, 1]$ in the lower limit topology. 
Case 1: 
If $x \in \left[ \frac{1}{2}, 1 \right]$, then 
  $$ 1 \geq x \geq \frac{1}{2},$$
  and so for any real number $y > x$, the open set $[x, y) \cap [0,1]$, for example,  contains $x$ but contains no point of $A$. 
Case 2:
If $x \in \left( 0, \frac{1}{2} \right)$, then 
  $$ 0 < x < \frac{1}{2}, $$
  and so 
  $$ 0 < \frac{1}{2} - x < \frac{1}{2}, $$
  and hence
  $$ \frac{1}{\frac{1}{2} - x } > 2.$$
  Thus there exists an $N \in \mathbb{N}$ such that 
  $$2 \leq N \leq \frac{1}{\frac{1}{2} - x } < N+1,  $$
  that is,
  $$ N+1 > \frac{1}{\frac{1}{2} - x } \geq N \geq 2, $$
  which implies that 
  $${1 \over {N+1}} <   {1 \over 2 } - x \leq  {1 \over N  } \leq { 1 \over 2 }.$$
  So 
  $$ 0 \leq {1 \over 2 } -  {1 \over N } \  \leq \  x  \ < \    {1 \over 2 } - {1 \over {N+1}}.$$
Thus if $x \neq \frac{1}{2} - \frac{1}{N}$, then the open set 
  $$ \left[ x,   {1 \over 2 } - {1 \over {N+1}} \right) = \left[x,   {1 \over 2 } - {1 \over {N+1}} \right) \cap [0, 1] $$ 
  contains  $x$ but this open set contains no point of $A$. 
And, if $x = \frac{1}{2} - \frac{1}{N}$, then $x \in A$ and thus the open set 
  $$ \left[ x,   {1 \over 2 } - {1 \over {N+1}} \right) = \left[x,   {1 \over 2 } - {1 \over {N+1}} \right) \cap [0, 1] $$ 
  contains  $x$  and  but this open set contains no point of $A$ other than $x$ itself. 
Case 3: If $x = 0$, then the open set $\left[0, \frac{1}{10} \right)$ contains only one point of $A$, namely the point $x = 0$ itself. 
Thus our set $A$ has no limit points in $[0, 1]$ as a subspace of $\mathbb{R}_l$.

Is the above reasoning correct? 
 A: Yes, it’s correct. 
You can also display an open cover with no finite subcover. One such is
$$ \mathscr{U} \colon= \left\{ \, \left[1-\frac1n,1-\frac1{n+1} \right) \, \colon \, n \in \Bbb Z^+ \,  \right\} \cup \big\{ \, \{ 1 \} \big\}.  $$
Each of these sets, except $\{ 1 \}$, is open in $\Bbb R_\ell$; $\{1\}$ is open in the subspace $[0,1]$, and $\mathscr{U}$ is a partition of $[0,1]$, so no proper subcollection of it covers $[0,1]$. 
We note that for $n = 1$, 
$$ \left[ 1 - \frac1n, 1-\frac1{n+1} \right) = \left[ 0, \frac12 \right), $$
which contains the point $0$.
Let $x \in (0, 1)$. Then 
$$ 0 < x < 1, $$
and so
$$ 0 < 1-x < 1, $$
which implies
$$ \frac1{1-x} > 1, $$ 
and so there exists an $N \in \mathbb{Z}^+$ such that $N \geq 2$ and 
$$ 1 \leq N-1  \leq  \frac1{1-x} < N, $$
which implies
$$ 1 \geq \frac1{N-1} \geq  1-x > \frac1N, $$
which in turn implies
$$ 0 \geq \frac1{N-1} - 1 \geq -x > \frac1N  - 1, $$
and hence
$$ 0 \leq 1 - \frac1{N-1} \leq x < 1 - \frac1N. $$
Let us put
$$ K \colon= N-1. $$
Then 
$$ x \in \left[ 1 - \frac1{K}, 1 - \frac1{K+1} \right). $$
