First examine the order on $\mathbb R \times \{ -1,0,1 \}$. Since this is lexicographic, we have that $$(x,i) \leq (y,j) \Leftrightarrow \begin{cases}
x < y, &\text{or} \\
x = y\mathrel{\&}i \leq j.
\end{cases}$$
In particular, $(x,-1) < (x,0) < (x,1)$ are three consecutive elements of the order for all $x \in \mathbb R$. Note, too, that if $(x,1) < (y,i)$ then it must be that $x < y$ and we also have $$(x,1) < (\tfrac{x+y}2 , -1 ) < (\tfrac{x+y}2 , 0 ) < (\tfrac{x+y}2 , 1 ) < (y,i).$$
- Using this it follows that $\{ (x,0) \} = ( (x,-1) , (x,1) )$ is an open set in $\mathbb R \times \{-1,0,1\}$ for all $x$, and so $\mathbb R \times \{ 0 \}$ is a discrete subspace (all subsets are open).
- Also, for any $x \in \mathbb R$ we have that sets of the form $$( (x,0) , (b,1) ) = \{ (y,i) : (x,0) < (y,i) < (x,b) \}$$ for $b > x$ form a basis for the open neighbourhoods of $(x,1)$. But since $(x,0) < (y,i)$ is the same as $(x,1) \leq (y,i)$, we can re-write these as "half-open intervals" $[ (x,1) , (b,1) )$. It is easy to see that $$[ (x,1) , (b,1) ) \cap ( \mathbb R \times \{ 1 \} ) = [ x , b ) \times \{ 1 \}$$ and it follows that the subspace $\mathbb R \times \{ 1 \}$ is homeomorphic to the "lower-limit topology" on $\mathbb R$.
To show that the subspace $\mathbb R \times \{0,1\}$ is Lindelöf we take advantage of the fact that the lower-limit topology is Lindelöf. It $\mathcal U$ is an open cover of $\mathbb R \times \{0,1\}$, then $\{ U \cap ( \mathbb R \times \{ 1 \} ) : U \in \mathcal U \}$ is an open cover of the Lindelöf $\mathbb R \times \{ 1 \}$, and so there is a countable subfamily $\{ U_n : U \in \mathcal U \}$ such that $\mathbb R \times \{ 1 \} \subseteq \bigcup_n U_n$. We now argue that $( \mathbb R \times \{ 0,1 \} ) \setminus \bigcup_n U_n$ is at most countable.
If it were uncountable, then the set $A = \{ (x,1) : (x,0) \notin \bigcup_n U_n \}$ would have a condensation point $(a,1)$ (actually, just having an accumulation point would do). Now $(a,1) \in U_n$ for some $n$, and so there is a $b > a$ such that $[ (a,1) , (b,1) ) \subseteq U_n$. But now uncountably many elements of $A$ belong to $[ (a,1) , (b,1) )$, and for each of these $(x,1)$ (except possibly $(a,1)$ itself) it follows that $(x,0) \in U_n$, contradicting that they are in not in $\bigcup_n U_n$!
As $( \mathbb R \times \{ 0,1 \} ) \setminus \bigcup_n U_n$ is countable, we just take countably many more sets from $\mathcal U$ to cover them.
