According to Rudin a point $p$ is said to be a limit point of a subset $E$ of a metric space $X$ if every neighborhood $N$ of $p$ contains a point $q \neq 0$ such that $q\in E$.
A point a point $p$ is said to be an interior point of a subset $E$ of a metric space $X$ if there exist a neighborhood $N$ of $p$ such that $N \subset E$.
Examples given shows that there may be limit points which are not interior points : $(a,b)$. There may be sets which does not contain any of its limit points and no point of it is an interior point.
Is interior points of a subset $E$ of a metric space $X$ is always a limit point of $E$ ? If $p$ is an interior point of $E$ then there exist a a neighborhood $N$ of radius $r$ such that $N \subset E$. Then any neighborhood of $p$ having radius less than $r$ will be contained in $N$ and hence contained in $E$. Any neighborhood of $p$ having radius greater than $r$ contains $N$ and hence intersect $E$ at some points other than $p$. Hence $p$ is a limit point of $E$. Is my argument is correct ?
Generally - In topology- this proof will not work as it does not have radius concept. But I want a counter example from metric space if I am wrong. Particularly a subset $E$ of $\mathbb{R}$ which has an interior point which is not a limit point of $E$.