Consider a set $E=[0..1] \cup \{10\}$
Note, that in (a) p may be equal to q.
So, according to (b) all points $[0..1]$ are limit.
Point 10 is isolated (c)
According to (b) every limit point of $E$ (which all are$[0..1]$) are points of $E$, so E is closed
All points of $E$ are interior(e), because all have neighborhood $\subset E$, even 10 itself, as $d(10,10) < r\ \forall r$
so by (f) $E$ is open.
What am I missing?
On
Note that $r>0$ . So 0 ,1and 10 are not interior points. Your argument for proving the closeness of the set is correct. I think you have an older edition of the book. Check the recent edition.
On
The fault is that $0,1,$ and $10$ are not interior points of $E$ because there exists no neighborhood of any of those points that is contained in $E$.
You can show this by supposing that the open ball of radius $r$ about $10$ is in $E$, and then noticing that this would imply $10+\epsilon$ is in $E$ for sufficiently small $\epsilon\ne 0$, which is false. Similar arguments apply for $0$ and $1$, by considering $0-\epsilon$ or $1+\epsilon$ with $\epsilon\gt 0$.
Thus, $E$ is not open. Your argument that $E$ is closed contains no fault, so $E$ is closed.
It is closed, but not open.
Not open: There is no neighborhood of the point $10$ that is contained in the set. Similarly for $0$ and $1$.
Indeed, let $r > 0$. We consider the neighborhood $N_r(10) = (10 - r, 10+r)$. It should be clear that $(10-r,10+r)$ is never entirely contained in $[0,1] \cup \{10\}$.
A set is open if all its points are interior points, or equivalently, if for all points in the set, there is a neighborhood of that point that is contained entirely in the set.
In symbols: $A$ open $\iff \forall a \in A: \exists r > 0: N_r(a) \subseteq A$
Negating this: $A$ is not open $\iff \exists a \in A : \forall r > 0: N_r(a) \not\subseteq A $
and I proved the latter statement.
Closed: Your argument here is correct. Alternatively, you can notice that the (finite) union of closed sets is closed, and a singelton and a closed interval are standard closed sets in metric spaces.
Addendum: (1) It is possible that a set is both open and closed (= clopen). For example, in any metric space $X$, $X$ and $\emptyset$ are clopen. If these two sets are the only clopen sets in the metric space, the metric space is called connected, as Rudin will define at the end of that chapter.
(2) A commonly made mistake is saying that $A \subseteq X$ is not open is the same thing as $A$ is closed. It is true however that $A$ is open is the same thing as $X \setminus A$ is closed. Don't confuse these two statements!