# Connectedness and limit points in $\mathbb{R}_l$

I am studying general topology and currently going through the sections on connectedness of Munkres' book.

I am studying the connectedness of $\mathbb{R}_l$. It is not connected as each interval $(-\infty,a)$ and $[b,\infty)$ is both open and closed. We can split each set by this logic until the only sets left are singletons, so those are the only connected sets.This means $\mathbb{R}_l$ is totally disconnected. However, Lemma 23.1 says:

If $Y$ is a subspace of $X$, a separation of $Y$ is a pair of disjoint nonempty sets $A$ and $B$ whose union is $Y$, neither of which contains a limit point of the other. The space $Y$ is connected if there exists no separation of $Y$

Then, if I take any separation of $\mathbb{R}_l$ e.g. $(-\infty,0) \cup [0,\infty)$. I see then that the second set contains a limit point of the first one, so by Lemma 23.1 $\mathbb{R}_l$ should be connected.

I would like to know why is not possible to use this Lemma on $\mathbb{R}_l$

In the topology on $\mathbb{R}_l$, neither $(- \infty, 0)$ or $[0, \infty)$ contains a limit point of the other. In particular, $0$ is not a limit point of $(- \infty, 0)$ because there is an open set containing $0$, namely $[0, \infty)$, that doesn't intersect $(- \infty, 0)$.
Sure, $0$ is a limit point of $(- \infty, 0)$ in the Euclidean topology on $\mathbb{R}$. But why should that matter? The Euclidean topology isn't the topology we are dealing with.
You are wrong: the second set contains no limit point of the first one. If $a\in[0,+\infty)$, then $[0,+\infty)$ is an open set to which $a$ belongs and $[0,+\infty)\cap(-\infty,0)=\emptyset$. So, $a$ is not a limit point of $(-\infty,0)$.