That point $a$ is not is not a limit point of $A$ precisely because it has a neighborhood, $W$, that does not contain any point of $A$ different from $a$ itself. (I’m assuming that you intended that $a$ belong to the set $A$, even though it’s detached from the rest of $A$.) Consider the set $A=(0,1]\cup\{2\}$. $1$ is a limit point of $A$, because every open set containing $1$ also contains other points of $A$. If $U$ is an open set containing $1$, then there is an $\epsilon>0$ such that $(1-\epsilon,1+\epsilon)\subseteq U$, and clearly
$$\max\left\{1-\frac{\epsilon}2,\frac12\right\}\in U\cap(A\setminus\{a\})\;.$$
$2$, on the other hand, is not a limit point of $A$, because the open set $(1,3)$ contains $2$ and no other point of $A$.
Finally, $0$ is a limit point of $A$, even though it does not belong to $A$: every open set $U$ containing $0$ contains an interval of the form $(-\epsilon,\epsilon)$, and
$$\min\left\{\frac{\epsilon}2,1\right\}\in U\cap(A\setminus\{0\})=U\cap A\;.$$
I) is the appropriate choice. If you want to be completely formal you might want to add a definition of what a neighborhood of $x$ is (in terms of the metric on the space).
II) does not work because it requires that $y$ belongs to every neighborhood of $x$. By the definition of a metric this can only be if $y=x$ which conflicts with the rest of the definition.
EDIT: (Drawing a picture in these cases can help quite a bit) Let $d$ denote the metric on $M$ and $B_r:=\{y\in M:d(x,y)<r\}$ denote the open ball centred around $x$ of radius $r$. Suppose that $y\in B_r$ for all $r>0$. Then, $d(x,y)=0$. Otherwise there exists an $r>0$ such that $d(x,y)=r$. However, this contradicts our assumption that $y
\in B_r$. Now, by the definition of the metric, $d(x,y)=0$ if and only if $x=y$.
Best Answer
Accumulation point $s$, of the set $T \subset \mathbb{R}$ or limit point $s$, of $T \subset \mathbb{R}$ are those points of $\mathbb{R}$ whose any nbd contains a point of $T$ other than $s$.
Thus for any nbd $S \in U(s)$ of $s$ if we shall delete $s$, then intersection of the deleted nbd $S-\{s\}$ with $T$ will be non empty.
So your expression $$(S -\{s\})\cap T \neq \phi$$ is correct.