Your proof is okay. More generally, we can define any topological space to be $T_1$ if its singleton sets are closed (or equivalently, for any points $x \neq y$, there is an open set containing $x$ and not containing $y$). $T_1$ is an example of a separation axiom. You've shown that metric spaces are $T_1$, but much stronger separation axioms hold for metric spaces.
Credit to Squirtle for providing similar details in a comment.
Your intuition in your first question is close, but we can be a bit more precise. Remember that a topological space is an ordered pair $(X,\mathcal{T})$, consisting of a set $X$ and a topology $\mathcal{T}$. A metric space on the other hand, is an ordered pair $(X,d)$, consisting of a set $X$ and a metric $d$. So, in that sense, a topological space and a metric space are fundamentally different things. That said, there is a natural way to get from a metric space to a topological space. Since metrics let us define open balls, we can use a metric to define a topology. In the case where the metric $d$ induces the same topology $\mathcal{T}$, we say that the topological space $(X,\mathcal{T})$ is metrizable. This is the case with $\mathbb{R}$ and the usual metric. The topology generated by open intervals, and the topology generated by open balls with the standard Euclidean metric, are the same, and so $\mathbb{R}$ with the open interval topology is metrizable. In this case it doesn't really matter whether we think of having a metric space or a topological space, but depending on what we're doing it may be useful to think of it one way or the other.
You may also be interested in reading about metrizable spaces, and examples of topological spaces which are not metrizable, for example $\mathbb{R}$ with the lower limit topology.
As for your second question, it's important to remember that open and closed are not the only types of sets. We say that a set is closed if its complement is open, but that's not the same as saying that a set which is not open is closed (which is false). For $[0,2)$, the complement is $(-\infty,0)\cup[2,\infty)$, which is not open, and so we know that $[0,2)$ is not closed. Because we also know that $[0,2)$ is not open, we conclude that it is neither open nor closed.
Best Answer
You define a metric space by $(X,d)$ where $X$ is a non-empty set and $d$ is the distance function. In the metric $(X,d),\,\, X$ is the universal set. So $X$ is always an open set. Now if you take $X$ as a singleton set then $X$ is always open.
Consider the Discrete metric space(trivial metric space) with $X=\mathbb Z$ or any subset of $\mathbb Z$. If you take any $0<r<1$ then every singleton set consisting a single integer is open in $X$.
In the usual metric, (Euclidean metric of degree 1) $(\mathbb R,d)$ no singleton set is open.