Why does this proof fail? I'm reading some notes on topology, and the notes' author is trying to raise motivation to consider compactness by providing a theorem whose proof is built intentionally wrong, but I don't agree with the reason he gives why the proof fails. Here's the theorem,
Every sequence $0<a_n<1$ has a subsequence $a_{n_k}$ with $0<\lim a_{n_k}<1$
and here's the (wrong) proof:
Let's look at  a collection of intervals of the form $$(a_n- \frac{1}{2n} ,a_n+ \frac{1}{2n})$$ with n=1,2... Each of these intervals has length,  1/n, and since $\sum_{i=1}^{\infty} 1/n =\infty$ and the length of the interval (0,1) is 1, there must be a point l in (0,1) which is covered by infinite intervals, say $(a_{n_k}-\frac{1}{2n_k},a_{n_k}+\frac{1}{2{n_k}} )$, therefore
$$\lim a_{n_k}=l\in(0,1)$$
The author says the proof is wrong because one can not always pick a finite subcover from an infinite cover (hence because the interval $(0,1)$ it's not compact), but I claim the proof breaks down even before that. 
Why? Because there isn't always a point in $(0,1)$ which is covered by infinite intervals, and an example of a collection of intervals for which there isn't a point $l$ in $(0,1)$ which is covered by infinite intervals is obtained when we substitute $a_n$ for $\frac{1}{2n}$, that is, the collection $\{(0,1/n), n\in \mathbb{N}\}$. 
So my question is, where does the proof actually fail? And, if the author was right, could you give a better explanation than his? 
Thank you for your help.
 A: You're exactly right about where the error is.  There are actually two different difficulties with concluding that there exists an $l\in(0,1)$ which is in infinitely many of the intervals $(a_n-1/2n,a_n+1/2n)$, though only one of them is fatal.  The first difficulty is that if $a_n$ is within $1/2n$ of $0$ or $1$, these intervals will not be entirely contained in $(0,1)$.  This, however, is easily remedied: just take some interval of length $1/n$ containing $a_n$ and contained in $(0,1)$ (though it may not be centered at $a_n$).  
The second difficulty is, as you observe, that even though the total length of these intervals is infinite, there need not be any single point that is in infinitely many of them.  All you can conclude is that there is no uniform bound $N\in\mathbb{N}$ such that every point of $(0,1)$ is in only $N$ of the intervals (because if there were, then the total length could be at most $N$).  But you might have points that are in arbitrarily large finite numbers of the intervals without having a point that is in infinitely many of them, and this is exactly what happens in your example.
It is entirely unclear what the author is trying to say; there is no obvious "open cover" here that you are taking a finite subcover of.  Moreover, this proof would also fail in $[0,1]$ (even though the conclusion is true in that case), because you can give the exact same counterexample to the argument.
A: I agree with you.  The erroneous assumption  $E$ is just as you say : $\{(a_n-1/2n,a_n+1/2n)\} $ may be a  point-finite family in $(0,1)$. Of course if you discard the assumption $E$ but erroneously assume that $(0,1)$ is compact, you will deduce immediately the erroneous result that any convergent subsequence of $a_n)_{n\in N}$ converges to a point in $(0,1)$ because any metric on a compact metrizable space is a complete metric.
