Understanding proof of, any compact subset of $\mathbb{R_l}$ must be atmost countable I was reading wiki proof of the result: "any compact subset of $\mathbb{R_l}$ must be at most countable" where  $\mathbb{R_l}$ means $\mathbb{R}$ with lower limit topology.
Proof is like this,  consider a non-empty compact subset $C⊆\mathbb{R_l}$. Fix an $x∈C$, consider the following open cover of $C$
$\{[x,∞)\} ∪\{(-∞,x-\frac{1}{n}) |n∈\mathbb{N}\}$
Since $C$ is compact, this cover has a finite subcover, and 
hence there exists a real number $a(x)$, such that the interval  $(a(x),x]$ contains no point of $C$ apart from $x$, This is true for all $x∈C$ 
Now choose a rational number, 
$q(x)∈(a(x), x]∩\mathbb{Q}$, Since the intervals $(a(x),x]$ parametrized by $x∈C$  are pairwise disjoint, the function $q:C→\mathbb{Q}$ is injective, and so $C$ is at most countable.
I didn't understand proof from "hence there exists(which is indicated in bold). Further how they conclude the function $q:C→\mathbb{Q}$ is injective and which function they are talking about? They did not defined any function!! 
If possible please explain me the proof from "hence there exists" OR is there any simple proof of above result? Please help me! "I do not want to "skip the proof". :-(
 A: Let $$U=\{[x,\infty)\} ∪\{(-\infty,x-\frac{1}{n}) |n∈\mathbb{N}\}$$ be your cover, and let $V\subset U$ be a finite cover of $C$. Then there are only finitely many integers $n$ such that $V$ contains $(-\infty,x-1/n)$, so let $N$ be the maximal such $n$. Then $C$ is covered by
$$\{[x,\infty),(-\infty,x-1/N)\}.$$ 
It follows that the intersection $$C\cap(x-1/N,x)$$ is empty. We then pick $q(x)\in\mathbb{Q}\cap(x-1/N,x)$. The point here is that for every $x$ we can pick an $N$ as described, then we let $a(x)=x-1/N$, and we pick a $q(x)\in(a(x),x)$. We make this choice for every $x\in C$, and this gives a function $q:C\to\mathbb{Q}$.
Now, to prove that $q$ is injective. Let $x,y\in C$ such that $x\ne y$ and $q(x)=q(y)$. Then $q(x)\in(a(x),x)\cap (a(y),y)$. If $y<a(x)$ then this intersection is very clearly empty, so we must have $y>a(x)$, but remember $(a(x),x)\cap C=\emptyset$, and $y\in C$, so we actually have $y>x$. But by a symmetrical argument we also have $x>y$, so by contradiction we have $q(x)\ne q(y)$, so $q$ is injective.  
