The set of all subsequential limits of a bounded sequence is a non-empty compact set Let $(x_n)$ be a bounded sequence and let $Y$ be the set of all subsequential limits of $(x_n)$. Prove that $Y$ is a non-empty compact set.
I think it's possible to solve this problem by proving that $Y$ is bounded (because if $Y$ is unbounded then $(x_n)$ is not bounded) and closed (because $\mathbb{R}^n-Y$ is open). I guess we can also use the definition of compact set by sequence (but in order to this, it's also necessary to prove that $Y$ is bounded).
However, I'd like to prove it by using cover. In other words, I'd like to prove that every open cover of $Y$ has a finite subcover (without using of Borel-Lebesgue Theorem, obviously). Is it possible?
Thanks.
 A: I suspect it is possible, although the resulting proof would probably be very similar to proving it using the definition of sequential compactness. The proof would probably go by contradiction: suppose you have an open cover $\mathcal C$ of $Y$ has no finite subcover. Let $y_1\in Y$, and choose $U_1 \in \mathcal C$ such that $y_1 \in U_1$. $U_1$ can't cover $Y$, so choose $y_2 \in Y\setminus U_1$, and choose $U_2 \in \mathcal C$ such that $y_2 \in U_2$. $U_1 \cup U_2$ can't cover $Y$, so choose $y_3 \in Y\setminus (U_1 \cup U_2)$, and choose $U_3 \in \mathcal C$ such that $y_3 \in U_3$. And so on. You should (I hope) be able to show that the resulting sequence $\{y_1,y_2,\dots,\}$ has no limit point, a contradiction.
Alternately, choose $y_j$ and $U_j$ as above, and then choose $x_{k_j}\in U_j$ (possible since $y_1$ is a subsequential limit); then the fact that $\{x_{k_j}\}$ should have a subsequential limit in $Y$ follows more directly from the definition of $Y$.
Such a proof might need to ensure that the sequence $\{U_j\}$ does form a cover of $Y$; you can arrange this by finding a countable subcover of $\mathcal C$ to start with, and then using the above construction for $U_{2j-1}$ and just exhausting the countable subcover set by set for $U_{2j}$. Or something like that....
