Why is the intersection of an open set and a single point open? I'm reading baby Rudin as independent study and he has the following theorem on page 36 (second edition):
Let $X$ be a metric space. Suppose $Y\subset X$. A subset $E$ of $Y$ is open relative to $Y$ if and only if $E = Y \cap G$ for some open subset $G$ of $X$.
My question is the following. Let's say I define
$Y = \{p\} \cup S$ where $S \subset X$, $p \in X$, $p \notin S$ and define $G$ such that $G \cap S = \emptyset$ and $p \in G$ and $G$ is open.
So $Y \cap G = E = \{p\}$
In what sense is $\{p\}$ open relative to $Y$? There is certainly no positive radius neighborhood of $p$ that is contained in $E$.
 A: Let's be a little more specific $X=\mathbb{R}$ with the usual metric. $S:=[0,1]$ and $p:=2713$. Then $Y=[0,1]\cup \{2713\}$. 
On $Y$ you have the induced metric $d_Y$ by the metric on $\mathbb{R}$. I claim that the ball of $(Y,d_Y)$ centered in $2713$ of radius $100$ is included in $\{2713\}$, hence $\{2713\}$ is open (in $Y$, of course). So there is no problem...
In general, the problem (I think this is the problem at least) is that you still consider open ball as open ball in $(X,d_X)$ but you really need to consider them as open ball in $(Y,d_Y)$ if you do so there is no contradiction. 
A: If $p$ is not limit point of $S$ then let $d(p,S)=r(\neq 0).$ Then $B_{r/2}(p)=\{p\},$ where $B_{r/2}(p)$ is open ball centred at $p$ of radius $r/2$ in $Y.$ Think what happens if $p$ is limit point of $S.$
A: Then $\{p\}$ is indeed an open subset of $Y$. 
Observe that $G$ is an open set with $p\in G$. 
For $r>0$ small enough we have $\{y\in X\mid d(y,p)<r\}\subseteq G$ so that $\{y\in X\mid d(y,p)<r\}\cap S=\varnothing$.
Consequently:
 $$\{y\in Y\mid d(y,p)<r\}=\{p\}$$
The LHS is an open ball in $Y$.
