For a topological space $(X,\tau)$, the topology $\tau$ on the set $X$ is a family of subsets called open sets, if $X$, $\emptyset$, any union of the subsets, and any finite intersection of the subsets are in $\tau$. This could be a definition of the term "open sets" with respect to topology.
On the real line, $X = \mathbb{R}$, we say $(0,1)$ is "open" based on the definition above, because $\tau_1 = \{\emptyset, \mathbb{R}, (0,1)\}$ can be a topology. However, we say $[0,1]$ is "not open", though $\tau_2 = \{\emptyset, \mathbb{R}, [0,1]\}$ can also be a topology that satisfies all of the properties. I think $[0,1]$ is not called open just because the $\tau_2$ topology is not so-called a standard topology of real line, correct?
I remember the definition of open set for a metric space $(X,d)$ is, we say a subset $S \subset X$ is open if $\forall x \in S$, $\exists \epsilon > 0$ s.t. $B_{\epsilon}(x):=\{y \in X \;|\; d(x,y) < \epsilon\} \subset S$. Is this exactly the standard topology of real line I am asking for, if we collect all of these subsets of $\mathbb{R}$? In other words, we can define open sets in such a way, and collect all these subsets of $\mathbb{R}$ to make a topology, called standard topology. Am I right?
One more question, it seems natural if we take the power set of $\mathbb{R}$, i.e., the discrete topology instead. Why is it not "standard"?