Prove that a metric space which contains a sequence with no convergent subsequence also contains an cover by open sets with no finite subcover. I really need help with this question:
Prove that a metric space which contains a sequence with no convergent subsequence also contains an cover by open sets with no finite subcover.
 A: Let $(a_n)$ be a sequence in the metric space $M$ that doesn't have any convergent subsequence. The set $\{a_n\}$ consists of isolated points (that is, it doesn't have any accumulation points; otherwise you could take a convergent subsequence), and it's infinite (because if it wasn't, one of the points would repeat infinitely in the sequence and we could get a constant subsequence). Now, for each $n$ take an open ball $B_{r_n}(a_n)$ around $a_n$ with radius $r_n$ small enough not to contain any other terms in the sequence, constructing an (infinite) family of open sets $\{B_{r_n}(a_n)\}$.
Now if you know some facts about compactness you could argue:
The set $\{a_n\}$ is closed, and then since it's a subset of the metric space $M$, if $M$ is compact then so is $\{a_n\}$. However, note that $\{B_{r_n}(a_n)\}$ is an open cover of $\{a_n\}$, but contains no finite subcover. Then $\{a_n\}$ isn't compact, and neither is $M$ (that is, not every subcover of $M$ admits a finite subcover).
Or otherwise, like suggested by Martin Sleziak (and showing a direct proof):
consider the open cover $\{B_{r_n}(a_n)\} \cup (M \setminus \{a_n\})$, which doesn't admit a finite subcover.
It's a nice exercise trying to fill in the details.
An example is given in the space of bounded sequences of real numbers, with the supremum norm: for $i \in \mathbb{N}$, let $e_i$ to be the sequence with a one at the $i$-th position and zeroes elsewhere; then $(e_i)$ is a sequence with no convergent subsequence and $\{e_i\}$ doesn't have accumulation points (even more: it's uniformly discrete).
