Subsequence of a sequence converging to its lim sup and lim inf $ Let (x_n)$  be a bounded real number sequence and $ (x_n )_{n≥k}  $be a subsequence of $x_n$  which only takes the values of the sequence starting from the k−th term. 
Let {$x_n $}  and {$x_n$ }$_{n≥k}$ denote a subset of R that contains all values of the corresponding sequences $(x_n )$  and $(x_n )_{n≥k}.$     
Define
            $α_k$= sup{$x_n$}$_{n≥k}$
               $\beta_k$= inf{$x_n$}$_{n≥k}$
              $x^∗$=lim sup⁡ $x_n$ = inf{$\alpha_k$}
 $x_*$=lim inf⁡ $x_n$ = sup{$\beta_k$}  
Using the above definitions, how do i prove that for every ( bounded? ) sequence  $(x_n)$ there is a subsequence that converges to lim sup $x_n$ and another subsequence that converges to lim inf $x_n$ ?
Also, what's the proof or reasoning of the fact that all subsequences of $(x_n)$ can only converge to values in the following interval [lim inf $x_n$ , lim sup $x_n$].     
After proving that, the important Bolzano-Weierstrass Theorem would follow imediately.      
I couldn't find any math resources which uses the lim inf, lim sup treatise ( not touching on topology ) and develops the proofs of many important theorems ( such as Squeeze theorem, Bolzano-Weierstrass, Limits and unequalities Theorem ) as corollaries of that .
Any recommendation ( a book, article, page, video, etc ) would be really helpful as well.
Thanks a lot in advance.  
 A: Let
$$x^* = \limsup x_n$$
and 
$$\alpha_n = \sup_{k\geq n}x_k$$
Since $x_n$ is bounded, $\alpha_n$ is bounded below and non-increasing.
Hence,
$$\lim_{n \rightarrow \infty}\sup_{k\geq n}x_k= \inf_{n}\sup_{k\geq n}x_k=x^*$$
Given $n \in \mathbf{N}$ there exists $m_n \geq n$ such that
$$x^* - 1/n < \alpha_{m_n} < x^* + 1/n.$$
Since $x^*-1/n < \sup_{k\geq m_n}x_k$,  there exists $k_n \geq m_n$ such that
$$x^* - 1/n < x_{k_m} \leq\alpha_{m_n}< x^* + 1/n.$$
Hence, $|x_{k_n} - x^*| < 1/n$ where $k_n \geq n$ and the subsequence $(x_{k_n}$) converges to $x^*$.
You can make a similar argument for $\liminf x_n$.
To show that no subsequence can converge to a value greater than $\limsup x_n$, assume that a subsequence converges to $x' > x^*$.  Let $\epsilon = (x'-x^*)/2$.  Then there are infinitely many $x_n$ greater than $x^* + \epsilon$, a contradiction of the basic property of $x^* = \limsup x_n$.
Again, try to make similar argument for $\liminf x_n$.
A: Hint: Can you show the following? For every $\epsilon > 0$, there are infinitely many $x_n$ satisfying $x_n > x^* - \epsilon$ but only finitely many such that $x_n > x^* + \epsilon$.
