Is it true that every singleton set in a Hausdorff space has no limit points? Here's my attempt at a proof:
Let $x \in X$ Hausdorff and consider the one-point set $\{x\}$. Let $y$ be a limit point of $\{x\}$. Then by def. every neighborhood of $y$ intersects $\{x\}$ at some point other than itself. If $y \in \{x\}$ then $y=x$, but then $X$ is a neighborhood of $y$ such that $X \cap \{x\} = \{x\} = \{y\}$, a contradiction. If $y \notin \{x\}$ then $x$ and $y$ are distinct, so there exist neighborhoods $U_{x}$ and $U_y$ such that $U_{x} \cap U_y = \emptyset$. Since $\{x\} \subset U_x$ it follows that $\{x\} \cap U_y = \emptyset$, a contradiction. Thus $\{x\}$ has no limit points.
Best Answer
Your answer is correct, although it can be generalized to $T_1$-spaces:
Suppose $x \in X$, where $X$ is a $T_1$-space. As you have shown, $x$ is not a limit point of {$x$}. On the other hand, if $y \neq x \in X$, we have that $X \setminus${x} is an open nhood of $y$ in $X$, which is disjoint from {x} (giving us $y$ is not a limit point of {$x$}).
It should be noted that it does not hold for $T_0$-spaces:
Consider $X =$ {$0,1$} with the Sierpiński topology (i.e., $\emptyset$, {$0$}, and $X$ are the only open sets). Then {$0$} has $1$ as a limit point.