If $x$ is a limit points in $\mathbb{R}_{\ell}$, is it necessarily a limit point in $\mathbb{R}$? Claim: $x$ a limit point in $\mathbb{R}_{\ell}$ $\Rightarrow$ $x$ a limit point in $\mathbb{R}$. Converse does not hold.
Given an element $[a,b)$ of $\mathbb{R}_{\ell}$ and $[a,b)$ of $\mathbb{R}$, the first does not have a limit point but the latter has $b$ as a limit point.  
However, given $(a,b) \in \mathbb{R}_{\ell}$, we have that $a$ is a limit point and also a limit point in the standard topology.
(I am trying to prove that $[0,1)$ has no limit points in $\mathbb{R}_{\ell}$, but how do I know that I only need to verify that $1$ is not a limit point.)  The claim above does not seem to cover this.  What am I missing?)
 A: I see some confusion in you: maybe this answer may help. I assume the following definition:

Given a topological space $X$, a subspace $A \subset X$ and a point $a \in X \setminus A$, we say that $a$ is a limit point of $A$ if every open neighbourhood of $a$ intersects $A$.

In particular, if a set $X$ has two topologies $\tau_1 \subset \tau_2$, then every limit point of $A$ in $(X,\tau_2)$ is a limit point of $A$ in $(X,\tau_1)$ as well (this is a generalization of your claim).
Suppose that $a \in \Bbb{R}_{\ell}$ is a limit point of $[0,1)$. Then, by the claim, $a$ is a limit point of of $[0,1)$ in the standard topology as well. Since the standard topology is well-known, it is quite easy to conlude that necessarily $a=1$. This means that, in order to recognize all limit points of $[0,1)$ in $\Bbb{R}_{\ell}$, it is sufficient to check whether or not $1$ is a limit point of $[0,1)$ in $\Bbb{R}_{\ell}$.
Now, we see that this is not the case, since the point $1$ has the open neighbourhood $[1,2)$, which is disjoint from $[0,1)$.
