Is a closed interval an open set? I read somewhere that a closed interval is not an open set but I don't see why not?
Some definitions in metric space: $(X,d)$
Open Ball: Let $p \in X$ and $r>0$ then $B(p,r) = \{ x \in X : d(x,p) < r\}$
Open set: A $S \subseteq X$ is open set if $\forall p \in S, \exists r>0$ such that $B(p,r) \subseteq S$
Now let's take $X= [a,b]$ and $d(x,y)= |x-y|$ then clearly an open ball can be made for $a<x<b$, for $x= a$, we see $B(a,r) = [a,r) \subseteq X$ if $0<r<b$ and similarly it works for $x=b$, so the closed interval is open set, yeah?

It depends on what you mean. If $X=[a,b]$ then the open sets (in the subspace topology) are of the form
$[a,b] \cap U$ where $U$ is open (in the usual sense) in $\mathbb{R}$. For example, $(a,b]$ is open.
If $X=\mathbb{R}$, then $[a,b]$ is not open because any open set that contains $b$ also contains points in $(b,\epsilon)$ for any $\epsilon>0$.
