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?
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 $$ x \in \left[ 1 - \frac1{K}, 1 - \frac1{K+1} \right). $$