Is this calculus proof I came up with sound? We want to prove there every bounded sequence has a converging subsequence. Let $[l,u]$ be the interval to which we know $a_n$ is bounded.Let $\{a_n\}$ be the sequence and  $[l_i,u_i]$ where $i$ is a positivie integer be a sequence of closed intervals such that each of the intervals contains an infinite amount of elements of ${a_n}$ and $[l_n,u_n]$ is always one of the two halfs of the sequence $[l_{n-1},u_{n-1}]$ and $[l_1,u_1]=[l,u]$ Pick a subsequence of $\{a_n\}$ called $\{b_n\}$ where $b_n\in[l_n,u_n]$. Then it converges since the sequence $(u-l)\frac{1}{2^n}$ converges, so for each $\epsilon>0$ there is an $N$ such that $u_N-l_N<\epsilon$. So if $N'>N$ then the distance from $b_{N'}$ to the midpoint of $[l_N,u_N]$ is less than $\epsilon$ since $B_{N'}\in[l_n,u_N]$.

For $i=1$ we set $l_1=l$, $u_1=u$ and $n_1=1$.
For $i>1$ we consider the interval $[l_{i-1},u_{i-1}]$ and split it into two intervals $\left[l_{i-1},\frac{l_{i-1}+u_{i-1}}{2}\right]$ and $\left[\frac{l_{i-1}+u_{i-1}}{2},u_{i-1}\right]$. Then in at least on of the two intervals there is an infinite number of elements of $\{a_n: n> n_{i-1}\}$. We set $l_i$ and $u_i$ accordingly and choose an index $n_i> n_{i-1}$ such that $a_{n_i}\in[l_i,u_i]$.
In this way we define a sub-sequence $b_i=a_{n_i}$ such that for all $j>i$ we have that $b_i$ and $b_j$ is contained in the interval $[l_i,u_i]$ and we conclude
$$|b_i-b_j| \leq (u_i-l_i) =\frac{u_{i-1}-l_{i-1}}{2} =\ldots = \frac{u-l}{2^{i-1}}.$$
Thus $\{b_i\}$ is a Cauchy sequence and therefore converges.
