Choosing a subsequence from the diagonal of subsequences. Selecting a subsequence from the diagonal of an array of subsequences is done in a hint for the solution of an exercise in "Introduction to Topology", by Gamelin and Greeene (Ex 8, p.25 and p.200).  I have not cited the entire exercise and hint, because only the selection of the diagonal sequence is my interest:
A compact and bounded metric space (X, d) has a family E of non-empty, closed subsets.  A metric is $\rho$ given for the family $E$.  As in the hint, I consider a sequence of subsets of the family, $\{E_n\}$ .  Because X is compact, its covering by balls $B(x, 1/k)$ has a finite sub-covering by balls $B(x_i, 1/k)$, where $i\,\epsilon\,\{1,N\}$. The hint prescribes the selecting of subsequences as follows, almost verbatim (p.200):
Set $E_{0j}=E_j$, and for each $k\ge1$ define a subsequence $\{E_{kj}\}^\infty_{j=1}$ of $\{{E_n}\}$ so that $\{E_{kj}\}$ is a subsequence of $\{E_{k-1,j}\}$, and such that for each $x_i$ , either $$B(x_i,1/k)\cap\{E_{kj}\}=\emptyset \; for\, all\,j \quad or\,else\quad B(x_i,1/k)\cap\{E_{kj}\}\neq\emptyset \; for\, all\,j. $$ 
My problem is in choosing the elements of the next subsequence, how am I to know for a particular $x_i$ whether there is a countable infinity of elements that  meet that ball or miss it?   If I knew there were an infinity of misses or hits or an infinity of hits and an infinity of misses, I could select a subsequence of one type or the other. Clearly, there is a countable infinity of hits or of misses and maybe of both.  How do I know what is the case? There may be only a finite number of hits or only a finite number of misses.  
The need to select the diagnonal subsequence from the procedure I copied from  Gamelin and Greene (op cit) is in the proof of ex. 8, as I mentioned above.  In that proof, for $x\,\epsilon\,F_k$ the authors select a particular ball $B(z,1/k)$ that meets $F_k$, as there must surely be.  Thus $B(z,1/k)\,\cap\,F_k\ne\emptyset$. Thus in the subsequence $\{E_{kj}\}$ and all its successors for $m\gt k$ ,  $\{E_{mj}\}$ meets $B(z,1/k)$.  In particular then, $B(z,1/k)\,\cap\,F_m\ne\emptyset$.  This is what is required for the rest of the proof in the hint. 
My conclusion is that in constructing the the array of subsequences from which to extract the diagonal subsequence, I would not know for one of the ball centers, $x_i$, whether or not its ball will meet a countable number of the closed subsets of the preceding subsequence.  For a point $x$ of a diagonal element $F_k$,   once such an array and diagonal sequence have been established , there has to be a covering ball, $B(z,1/k)$ by the compactness of (X,d).  I find the argument troubling, because the array is established by "construction".  Yet, I do not see how the construction is to be made. 
 A: What you need for that step is the following lemma.

Lemma. Let $\mathscr{U}$ be a finite open cover of a space $X$, and let $\sigma=\langle E_n:n\in\Bbb N\rangle$ be a sequence of closed subsets of $X$. Then $\sigma$ has a subsequence $\langle E_{n_k}:k\in\Bbb N\rangle$ such that for each $U\in\mathscr{U}$, either $U\cap E_{n_k}=\varnothing$ for each $k\in\Bbb N$, or $U\cap E_{n_k}\ne\varnothing$ for each $k\in\Bbb N$.

To prove this, let $\mathscr{U}=\{U_1,\ldots,U_m\}$, say. We first find a subsequence of $\sigma$ that ‘works’ for $U_1$. Let $M=\{n\in\Bbb N:U\cap E_n=\varnothing\}$; if $M$ is infinite, let $\sigma_1=\langle E_n:n\in M\rangle$. Otherwise $\Bbb N\setminus M$ is infinite, and we let $\sigma_1=\langle E_n:n\in\Bbb N\setminus M\rangle$. In either case $\sigma_1$ is an infinite subsequence of $\sigma$; in the first case each term of it is disjoint from $U_1$, and in the second case each term intersects $U_1$.
Now replace $\sigma$ by $\sigma_1$, and use the same argument to find an infinite subsequence $\sigma_2$ of $\sigma_1$ such that either each term of $\sigma_2$ is disjoint from $U_2$, or each term of $\sigma_2$ intersects $U_2$. Repeat this until you have $\sigma_m$, an infinite subsequence of $\sigma$ with the desired property.
I sketched the argument pretty informally; if you want to do it more rigorously, you can cast it as an induction on $|\mathscr{U}|$. The argument in the first paragraph of the proof sketch is basically the induction step.
