Relation between connected sets and limit point I'm studing topology by Munkres book and I am intrigued with two results:

Lemma 23.1 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 whitch contains a limit point of the other. The space $Y$ is connected if there exists no separation of $Y$.
Theorem 23.4. Let $A$ be a connected subspace of $X$. If $A \subset B \subset \overline{A}$, then $B$ is also connected.
Said differently: If $B$ is formed by adjoining to the connected subspace $A$ some or all of its limit points, then $B$ is connected.

So, my question is: what's the interpetration or intuitive meaning and relation between limits point and connected spaces?
In their proofs, this conditions of limit points seems to me to be just a "technical condition". I try to look for some interpretation for these two results, but I can't come up with anything "palpable".
In the example 2. The subset $Y = [-1,0) \cup (0,1]$ of the real line $\mathbb{R}$, $[-1,0) $ and $(0,1]$ are open subsets of $Y$ and they form a separation of $Y$. In the standard topology of $\mathbb{R}$, $0$ is a limit point of booth sets, but neither of these sets contains $0$. If I "add" 0 in some of these sets, for example ${0} \cup (0,1]$ then, $[-1,0) \cup [0,1]$ is not equal to $Y$, therefore, is not a separation of $Y$.
 A: A separation of a topological space $Y$ is a pair of two disjoint open subsets $A$ and $B$ such that $A \cup B = Y$ and $A \cap B = \emptyset$. If a separation exists, $Y$ is disconnected.
We can prove the equivalence of this definition of separation with the one you reported in the lemma in the following way.
A subset $Y$ of a topological space $X$ can be regarded as a space itself, with the subset topology inherited by $X$. A subset $S \in Y$ is open in $Y$ if there exists an open set $P \in X$ and $S = P \cap Y$.
Let's take a pair of two disjoint non empty subsets $A$ and $B$ such that $A \cup B = Y$ neither of which contains an accumulation point of the other. An accumulation point $p \in X$ of $A$ is a point $p \notin A$ such that for each open  subset $V \in X$ we have $A \cap V \neq \emptyset$. As $B$ does not contain accumulation points of $A$, for each $b \in B$ there exists an open subset $V_b \in X$ which does not contain points of $A$. The union $\bigcup\limits_{b \in B} V_b$ is an open subset of $X$ and $B$ is a subset of $Y$ such that $ B = \bigcup\limits_{b \in B} V_b \cap Y$, hence $B$ is open in $Y$. Similarly, $A= \bigcup\limits_{a \in A} V_a \cap Y$ is open in $Y$, hence $Y$ as a space has a separation in two sets $A$ and $B$ that are open under the subset topology.
Conversely, if $A$ and $B$ are open in $Y$  and form a separation of $Y$, it means that for each $a \in A$ there exists an open subset of $Y$ $V_a \subset A$. As $A$ and $B$ are disjoint, for each $a \in A$ $a$ is not an accumulation point of $B$. Same for $A$.
I am not sure why $Y$ is indicated as a subset and not a space itself, but I think all I wrote above holds if we consider $Y$ as a space itself, without resorting to the subset topology. Note that $A$ and $B$ can be either regarded as subsets of $X$ or of $Y$.
Informally, if $A$ does not have accumulation points of $B$ means that all its points are "internal" ($B$ is the complement of $A$) hence it is an open set .
Hope this helps!
