Prove that K, a subset of R, is compact I thought this question is quite easy.
My proof is that
In R, the subset is compact if and only if is closed and bounded.
K is bounded since d(1/2, r) < 1 for any r in K, and
it is closed since its only limit point is 0 and it is in K.
Therefore K is compact.
However, the solution given by the MIT Open Course is much more complex and
I can't understand what this is about.
Is my solution valid? Can you help me understand the MIT Open course solution? 
I can't understand the part beginning from "Because epsilon > 0, there exists an N"
Problem:


MIT Opencourse Solution: 


 A: The idea is as follows. Let the open sets $O_\alpha\;,\;\alpha\in A$ cover your set. Then $0$ must be in some of them, call it $O_\beta$. But since $$\tag 1 \lim_{n\to\infty}\frac 1 n=0$$ and $O_\beta$ is open, there must exist $N$ such that $1/n\in O_\beta$ for $n=N,N+1,\ldots$. This stems directly from the definition of $(1)$ being true: given $\epsilon >0$, pick $N$ big enugh so that $1/N<\epsilon$. Then $$\frac 1n\in (-\epsilon,\epsilon)=B(0,\epsilon)\; \;,\; n=N,N+1,\ldots$$ Now that we took care of the infinitely many numbers $\geq N$, only a finite amount remain, and they can be put into a finite number of $O_\alpha$s, yielding a finite cover.
Maybe this picture will help:

A: Compactness is usually defined via open covers (i.e. a space is compact iff every open cover has a finite subcover).  So you have to prove it directly from this definition without relying on Heine-Borel.  
This problem is asking you to prove a special case of the following:
Let $X$ be a topological space and let $\{x_n\}_{n = 1}^\infty$ be a sequence in $X$ that converges to $x$.  Then the set $A = \{x_n : n \in \mathbb{N}\} \cup \{x\}$ is compact.
The proof proceeds along the same lines.  Given an open cover of $A$, it will have an open set containing $x$, but since $x_n \to x$ it will contain all but finitely many elements of $A$.  So we can choose finitely many open sets in the cover to take care of the remaining elements.   
In your case, just notice that $1/n \to 0$.   
