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 $B_{N'}\in[l_n,u_N]$.
 A: The proof works except for the last line, where there is some mess with the indexes $n,N,N'$. Moreover, why are you taking into account the midpoint of $[l_n,u_n]$? If I'm guessing correctly, you would like to prove that the sequence $(b_n)_n$ converges to $[l_N,b_N]$, but this is false (indeed, your $N$ depends on $\epsilon$). 
Since you cannot know exactly to which point $(b_n)_n$ converges to, you cannot prove its convergence using simply the definition.
You should use some convergence criterion.
Finally, you should be more precise in your statement by saying that you are considering sequences of real numbers.
A: 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.
