Let $X$ be a topological space and $A \subset X$. We say $x \in X$ is a ${\bf limit \; point}$ of $A$ if every neighborhood of $x$ intersects $A$ in a point different from $x$.
This can be rephrased as follows: Let $L$ be set of limit points of $A$. $x \in L$ means that for every $U$ open in $X$, $(U \setminus \{ x \} ) \cap A \neq \varnothing$.
Consider now $X$ with discrete topology. That is, let $\tau_X = \mathcal{P}(X)$. For simplicity, say $X = \{a,b,c\}$. For example, consider $A = \{b,c\}$. Observe that this has no limit point as the open $\{ a\}$ fails to intersect $\{b,c\}$ and the other points intersect $A$ on those points violating definition of limit point. Therefore, every subset of $X$ has no limit point... Does this generalize?
Another example, Consider $\mathbb{R}$ with standard topology and let $A = [0,1)$. Isnt every point of $A$ here a limit point of $A$ ? Indeed, take any $ 0 \leq x < 1 $ and choose some $U$ open so that $x \in U $ and clearly $U \cap [0,1) = \varnothing $.
Am I understanding these concepts correctly ?
It is indeed true that in any discrete space $(X, \mathcal{T}=\mathscr{P}(X))$ no subset $A$ of $X$ has a limit point, as for any $x$, $U=\{x\}$ is open and $U\setminus\{x\} = \emptyset$. So in that sense your observation generalises.
Note that the definition of limit point is "for every open set $U$ that contains $x$, we have that $(U\setminus \{x\}) \cap A \neq \emptyset$. (Otherwise only dense sets can have limit points, it's about the points "close to $x$")
In $\Bbb R$ in the standard topology things are different. $A=[0,1)$ has all points of $[0,1]$ as its limit points. If $x \in [0,1]$ and $U$ is open containing $x$, we know $(x-\delta, x+\delta) \subseteq U$ and we always have points of $[0,1)$ to the left or right of $x$ (tto the right only if $x=0$, and to the left only if $x=1$, but at least one of $x-\frac{\delta}{2}$ or $x+\frac{\delta}{2}$ is a point in $(U\setminus \{x\}) \cap [0,1)$ if $x\in [0,1]$. If $x>1$, $U=(1, x+1)$ is an open set that contains $x$ and misses $A$ entirely. Same for $U=(x-1,0)$ if $x<0$ so points outside $[0,1]$ are not limit points of $[0,1)$.
Now check that in the same topology $\Bbb Z$ has no limit points at all.