In $\mathbb{R}$ with the discrete metric, prove that $\{x\in \mathbb{R} : d(x,0)\leq 3\}$ is bounded and closed, however it is not compact. I know that with the discrete metric $d(x,0) = 0$ if $x=0$ and $d(x,0) = 1$ if $x\neq0$.
For the bounded we have that $r=3>0$ and I'm confused if I need to use the discrete metric since my only bounded definition so far is $A \subset E \iff (\exists r>0) (\exists x\in E) : A\subset B_r(x)$ but is that only for sets or does that apply to a metric space as well?
I also know that in a discrete metric everything is open so everything is closed but I am having formulating that into a proof. Is that because the options are $0$ or $1$ so $\mathbb{R}\in d(x,0)\leq 3$?
Finally I know there exists an infinite open cover $\{B_n(x)\}_{x\in\mathbb{R}}$ but I'm having trouble showing that as well.
Any help is appreciated, thanks!
 A: Answers, in short


*

*The definition for bounded sets applies to any metric space. In fact, there is a definition for bounded sets that extends to a larger class of spaces called as topological vector spaces.

*In your second paragraph, your logic is right i.e. $d(x,0) \leq 1$ for all $x$, so a ball around $0$ of radius $3$ covers every point of $\mathbb R$.

*There is an infinite cover with no subcover, and the answer to that question is about understanding the topology of our metric space.
Boundedness
It is obvious that the set $\{x \in \mathbb R : d(x,0) \leq 3\}$ is bounded, by taking $x= 0$ and $r = 3$ in the definition of boundedness.
Also, notice from my second bullet point that the above set is equal to $\mathbb R$, because every point of $\mathbb R$ is at distance less than or equal to $1$ (and therefore $3$) from $0$.
Closure
We have already seen that the above set is $\mathbb R$. Of course, $\mathbb R$ is closed, since its complement, the empty set, is open by definition (which is the usual definition of closed).
Furthermore, note that for every point $y \in \mathbb R$, we have that $\{y\} = B(y,\frac 12)$, because if $d(x,y) < \frac 12$ then $x = y$ must happen.
Consequently, $\{y\}$ is open for all $y$, and since arbitrary unions of open sets are open, every subset of $\mathbb R$ is open (and closed) under this metric.
Compactness
Now that we know all open sets of $\mathbb R$ under the discrete topology, it is easy to describe a cover that has no finite subcover. Indeed, just take the cover of singletons : $\mathcal U = \bigg\{\{y\} : y \in \mathbb R\bigg\}$. 
This is an open cover : it covers $\mathbb R$ obviously, and each set is open.
It has no finite subcover, obviously because if you take a finite subset $\mathcal U$, it will cover only finitely many points, because each open set has only one point. On the other hand, $\mathbb R$ is not finite, so you cannot cover $\mathbb R$ with this finite set.
Consequently, $\mathbb R$ is non-compact under the discrete metric.
Closed and bounded does not imply compact
In any metric space, if a set is compact, then it is definitely bounded and closed. The converse, as you have seen above, is NOT true.
It turns that a set is compact if and only if it is closed, and totally bounded. 

In a metric space, a set is totally bounded if for all $\epsilon > 0$, there exist points $x_1,x_2,...,x_n$ such that the balls $B(x_1,\epsilon), B(x_2,\epsilon),...,B(x_n,\epsilon)$ form a cover of the set.

You can check that $\mathbb R$ is not totally bounded under this metric by taking $\epsilon = \frac 12$, by the fact that $B(y , \frac 12) = \{y\}$ for all $y$, and $\mathbb R$ is infinite.
