Why isn't [0,1] an open set by this definition of Ross Defintion : Let $(S,d)$ be a metric space. Let $E$ be a subset of $S$. An element $s_0 \in E$ is interior to $E$ if for some $r > 0$ we have
$\{s \in S : d(s,s_0) < r\} \subseteq E$.
We write $E^\circ$ for the set of points in $E$ that are interior to $E$. The
set $E$ is open in $S$ if every point in $E$ is interior to $E$, i.e., if $E = E^\circ$.
By definition , I can make my $r$ very large, say $r=1000$ and have every point in $[0,1]$ has to have a distance of less than $1000$, so every point in my set is interior , thus $[0,1]$ is indeed an open set, but the textbook says that $[0,1]$ is a closed set. Could someone point out why my definition is wrong?
Pardon my English, it is my second language, if anything is unclear I can clarify it.
 A: Let's denote
$$B_\epsilon(s_0) = \{s \in S \ | \ d(s,s_0) < r\}.$$ These are called open balls, and to be a point in the interior of a subset $E$ in a space $S$ with a metric (or generally any topological space) there must exist an open neighborhood $U$ containing that point that is wholy contained within the subset $E$. In the context of this example, there must exist some $\epsilon >0$ such that $U=B_\epsilon(s_0) \subseteq E$. Note in the example you gave with $\epsilon = 1000$, the open ball fails to be contained within the set $[0,1]$.
Intuitively, you should think about points in the interior of the set $E$ as having "room around them," which is what this definition is precisely trying to say. You say that $s_0 \in E$ is in the interior if $s_0$ has some wiggle room:

But in contrast, if we considered some $s_0$ on the boundary, i.e. the black border above (or, in the case of $E = [0,1]$, the points $0$ or $1$), observe that they don't have any wiggle room! You can't find any $\epsilon > 0$ such that you can fit $B_\epsilon(s_0)$ inside $E$.
This is all considering that $E \neq S$! In other words, $(S,d) = (\mathbb{R},|\cdot |)$.
A: Others have explained why the proof you propose is wrong. There are still a few misconceptions in your post that I think are worth clearing up.
First is the notion of what an open set is. The definition you wrote is completely correct, but saying "$[0, 1]$ is an open set" is, to be pedantic, unclear. Being open is not a property of a set, it's a property of a subset of a topological space (in this case, lets just say of a metric space). It is not an absolute notion, but a relative one. Indeed, $([0, 1], |\cdot|)$ is itself a metric space, so $[0, 1]$ is an open subset of this metric space. However, it is not an open subset of the metric space $(\mathbb R, |\cdot|)$.
Also, that $[0, 1]$ is a closed subset of $\mathbb R$ is not a priori enough to conclude that it is not open. Although it is true (by connectedness) that proper nonempty closed subsets of $\mathbb R$ are not open, this is not a general fact about metric spaces. Being closed and being open are not mutually exclusive possibilities. For example, take $(0, 1) \cup (2, 3)$ with the restriction of the Euclidean metric. Then both subsets $(0, 1)$ and $(2, 3)$ are open, so $(0, 1)$ is both open and closed in this space.
