A topological space $X$ is said to be scattered if for every closed subset $C$ of $X$ the set of isolated points of $C$ is dense in $C$ Equivalently, $X$ is a scattered space if no non-empty closed subset of $X$ is dense in itself: for every closed subset $C$ of $X$ the closure of the interior of $C$ is not $C$

A subset of a topological space is called scattered if it is a scattered space with the subspace topology.

Every discrete space is scattered, since every singleton is open, hence isolated.

Scattered line. Let $\mathbb{R}$ be the real line equipped with the usual topology $T$ (formed by the open intervals). Let's define a new topology $S$ on $\mathbb{R}$ as follows: a subset $A$ is open under $S$ ($A\in S$ if $A=B\cup C$ where $B$ is open under $T$ ($B\in T$ and $C\subseteq \mathbb{R}-\mathbb{Q}$ a subset of the irrational numbers. We make the following observations:

1. $S$ is a topology on $\mathbb{R}$ which is finer than $T$
2. $\mathbb{R}$ is a Hausdorff space under $S$
3. a singleton in $\mathbb{R}$ is clopen iff it contains an irrational number
4. any subset of irrationals is scattered under the subspace topology of $\mathbb{R}$ under $S$

Proof.

1. First note that every element of $T$ is an element of $S$ so $\varnothing, \mathbb{R}\in S$ in particular. Suppose $A_1,A_2\in S$ with $A_1=B_1\cup C_1$ and $A_2=B_2\cup C_2$ where $B_i,C_i$ are defined as in the setup above. Then $A_1\cap A_2= B\cup C$ where $B=B_1\cap B_2\in T$ and $C=(C_1\cap B_2)\cup ((B_1\cup C_1)\cap C_2)$ is a subset of the irrationals. So $A_1\cap A_2\in S$ If $A_i\in S$ with $A_i=B_i\cup C_i$ then $\bigcup A_i=\bigcup B_i\cup \bigcup C_i\in S$ So $S$ is a topology which is finer than $T$
2. $\mathbb{R}$ is Hausdorff under $S$ is clear, the topological property is inherited from $T$
3. First, any singleton is closed since $X$ is Hausdorff under $S$ If $x$ is irrational, then $\lbrace x\rbrace$ is open (under $S$ as well. So $\lbrace x\rbrace$ is clopen. If $x$ is rational and $\lbrace x\rbrace\in S$ then it is the union of a $T$ open set $B$ and a subset $C$ of the irrationals. The only $T$ open subset of $\lbrace x\rbrace$ is the empty set, so $\lbrace x\rbrace$ is a subset of the irrationals, a contradiction.
4. Let $C$ is a subset of the irrational numbers. and considered the subspace topology under $S$ Then every point $r$ of $C$ is isolated, since $\lbrace r\rbrace$ is the open subset of $C$ separating it from the rest. The closure of the collection of these points is clearly $C$ itself, so $C$ is scattered.

Remark. Every topological space is a disjoint union of a perfect set and a scattered set.