Below is my attempt to prove the topological version of the Bolzano-Weierstrass Theorem. Is it an effective proof? I'd appreciate any comments on it. The book gave a hint to use a nested sequence of half-intervals. The idea is pretty intuitive...I'll explain it if anyone would like me to.
Bolzano-Weierstrass Theorem: "Every bounded, infinite subset of $\mathbb{R}$ has a limit point."
"Let $A$ be a bounded, infinite subset of $\mathbb{R}$. Then since $A$ is bounded, it is a subset of some closed interval $[a,b]$. Take a sequence of half-intervals of $[a,b]$, $\{[a_n,b_n]\}_{n=1}^\infty$ where $[a_1,b_1]=[a,b]$. By Cantor's Nested Intervals Theorem $\displaystyle\bigcap_{n=1}^\infty [a_n,b_n]$ is nonempty and since $\mbox{diam}([a_n,b_n]) \to 0$ as $n \to \infty$, the intersection contains exactly one element, say $p$. Since $p \in \displaystyle\bigcap_{n=1}^\infty [a_n,b_n]$ and every $[a_n,b_n]$ contains infinitely many elements of $A$*, so does $(a_n,b_n)$. Since $\mbox{diam}(a_n,b_n) \to 0$ as $n \to \infty$, every open set containing $p$ also contains some $(a_n,b_n)$, so $p$ is a limit point of $A$."
*Intuitively, if we cram an infinite number of points into a bounded interval, they will be 'dense' in that interval (I haven't formally learned what 'dense' means yet), but how do we prove it?