Cantor Intersection Theorem: If $X$ is a non-empty complete metric space, and $C_n$ is a sequence of closed nested subsets of $X$ whose diameters tend to zero: $$\lim_{n\to\infty} \operatorname{diam}(C_n) = 0$$ then the intersection of the $C_n$ contains exactly one point: $$\bigcap_{n=1}^\infty C_n = \{x\}$$ for some $x \in X$.
Bolzano–Weierstrass theorem: Each bounded sequence in $\mathbb{R}^m$ has a convergent subsequence.
Is the following proof of BW right?
Proof: We are given that there exists $M$ such that $\Vert x_k \Vert_m \leq M$ for all $k \in \mathbb{N}$. Now consider $C_1$ to be the box $\Vert x \Vert_m \leq M$ and define $C_{k+1}$ to be one of the $2^m$ disjoint sub-boxes of $C_k$ consisting of infinite $x_k$'s. Hence, we have $C_{k+1} \subseteq C_k$ and $$\text{diam}(C_{k+1}) = \dfrac{\text{diam}(C_{k+1})}2$$ Hence, using Cantor's theorem, we have that there exists $x$ such that $\displaystyle\bigcap_{k=1}^{\infty} C_k \in \{x\}$. Picking $x_k \in C_k$, we obtain the desired convergent sub-sequence.