I was reading Mathematical Analysis by Tom M. Apostol. There Cantor Intersection Theorem was proven using Bolzano-Weierstrass Theorem in this way
Theorem : Let $\left\{Q_{1}, Q_{2}, \ldots\right\}$ be a countable collection of nonempty sets in $\mathbb{R}^n$ such that:
i) $Q_{k+1} \subseteq Q_{k} \quad(k=1,2,3, \ldots)$
ii) Each set $Q_{k}$ is closed and $Q_{1}$ is bounded.
Then the intersection $\bigcap\limits_{k=1}^{\infty} Q_{k}$ is closed and nonempty.
Proof : Let $S=\bigcap\limits_{k=1}^{\infty} Q_{k} .$ Then $S$ is closed [As intersection of an arbitrary collection of close sets is closed]. To show that $S$ is nonempty, we exhibit a point $x$ in $S .$ We can assume that each $Q_{k}$ contains infinitely many points; otherwise the proof is trivial. Now form a collection of distinct points $A=\left\{{x}_{1}, {x}_{2}, \ldots \right\},$ where ${x}_{k} \in Q_{k}, $ since $A$ is an infinite set contained in the bounded set $Q_{1},$ it has an accumulation point, say $x$. We shall show that $x \in S$ by verifying that $x \in Q_{k}$ for each $k .$ It will suffice to show that $x$ is an accumulation point of each $Q_{k},$ since they are all closed sets. But every neighborhood of ${x}$ contains infinitely many points of $A,$ and since all except (possibly) a finite number of the points of $A$ belong to $Q_{k},$ this neighborhood also contains infinitely many points of $Q_{k}$. Therefore $x$ is an accumulation point of $Q_{k}$ and the theorem is proved.
Honestly I didn't understand the proof from the line
It will suffice to show that $x$ is an accumulation point of each $Q_{k},$ since they are all closed sets.
Can you help me to understand it.
Best Answer
If $x$ is an accumulation of points for each $Q_k$ then $x \in Q_k$ for all $k$ because $Q_k$ is closed. Hence $x \in \bigcap Q_k$ which is what we are trying to prove.
Now let $U$ be an open set containing $x$. If $U$ contains only finitely many points of some $Q_k$ then is contains only finitely many points of $\bigcup_{ i \geq k} Q_i$ because $Q_i \subseteq Q_k$ for all $ i \geq k$. It follows that $U$ contains only finitely many points of $A$. [Recall that we have picked just one point $x_k$ from each $Q_k$]. But this contradicts the fact that $x$ is an accumulation point of $A$. This complete the proof.