Why is $\mathfrak{c} = |\mathbb R|$ the cardinality of the lower limit topology on $\mathbb{R}$?
An open set in the lower limit topology is of the form $[a,b)$. I can clearly see why the cardinality has to be at least $\mathfrak{c}$, but why is it bounded by $\mathfrak{c}$?
There are open sets in the lower limit topology that are not of the form $[a,b)$: the lower limit topology consists of the subsets of $\Bbb R$ that are unions of sets of the form $[a,b)$, so, for instance, every set of the form $(a,b)$ is open in the lower limit topology:
$$(a,b)=\bigcup_{a<x<b}[x,b)\;.$$
Let $\tau$ be the lower limit topology on $\Bbb R$; there are several ways to see that $|\tau|\le\mathfrak{c}$. One is to observe that $\langle\Bbb R,\tau\rangle$ is hereditarily Lindelöf. Now let $U\in\tau$, and let $\mathscr{U}=\{[a,b):[a,b)\subseteq U\}$; clearly $\mathscr{U}$ is an open cover of $U$, so it has a countable subcover $\mathscr{U}_0$. Thus, every open set in the lower limit topology is the union of countably many sets of the form $[a,b)$. There are $\mathfrak{c}$ such sets, so there are $\mathfrak{c}^{\aleph_0}=(2^{\aleph_0})^{\aleph_0}=2^{\aleph_0\cdot\aleph_0}=2^{\aleph_0}=\mathfrak{c}$ countable families of them and hence at most $\mathfrak{c}$ members of $\tau$.
Here is another way to see this that takes advantage of a connection between closed sets in the lower-limit topology (lower-limit-closed sets), and closed sets in the usual/metric topology (metric-closed sets) which is similar to a step in this answer. Let $\operatorname{cl}_{\text{met}}$ denote the usual/metric closure operator.
Now, if $F \subseteq \mathbb R$ is lower-limit-closed, then clearly $F \subseteq \operatorname{cl}_{\text{met}} (F)$ (and since the lower-limit topology is strictly finer, this can be a strict inequality). I claim that $\operatorname{cl}_{\text{met}} (F) \setminus F$ is at most countable.
proof. Note that for each $x \in \operatorname{cl}_{\text{met}} (F) \setminus F$, as $\mathbb R \setminus F$ is an lower-limit-open neighborhood of $x$ there must be a $b_x > x$ such that $[ x , b_x ) \cap F = \emptyset$.
Note that if $x < y$ are both in $\operatorname{cl}_{\text{met}} (F) \setminus F$ and $[x,b_x ) \cap [y,b_y) \neq \emptyset$, then $x < y < b_x$. As $y \in \operatorname{cl}_{\text{met}} (F)$ but $y \notin F$ there must be a sequence in $F$ converging to $y$, but as $[y,b_y) \cap F = \emptyset$ it follows that this sequence can be chosen to be strictly increasing. Thus some tail of the sequence must be included in $[x,y) \subseteq [x,b_x)$, which contradicts our choice of $b_x$!
Therefore $\{ [x,b_x) : x \in \operatorname{cl}_{\text{met}} (F) \setminus F \}$ is a pairwise disjoint family of half-open intervals, and since each (non-degenerate) half-open interval must contain a rational number it follows that there can be only countably many.
It follows that every lower-limit-closed set can be obtained from a metric-closed set (its metric-closure) by removing at most countably many points. (Of course, not all such sets are lower-limit-closed, but this will provide an upper bound.) Now given a metric-closed set $E$, as $| E | \leq \mathfrak c$ it follows that at most $| [E]^{\leq \aleph_0} | \leq \mathfrak c ^{\aleph_0} = \mathfrak c$ many lower-limit-closed sets can be obtained from $E$ by removing at most countably many points. As there are $\mathfrak c$ many metric-closed sets it follows that there are at most $\mathfrak c \cdot \mathfrak c = \mathfrak c$ lower-limit-closed sets.
