Suppose that $(b_{n})$ is a fixed, bounded infinite sequence and \begin{align*} x_{n}&=\inf\{b_{n},b_{n+1},b_{n+2},\cdots\}\\ y_{n}&=\sup\{b_{n},b_{n+1},b_{n+2},\cdots\}\\ x&=\sup\{x_{n}:n\in\mathbb{N}^{+} \}\\ y&=\inf\{y_{n}:n\in\mathbb{N}^{+} \} \end{align*} Prove there is a subsequence of $ (b_{n})\ $that converges to $y$.
My attempt:
Since $ (y_{n})\ $is monotone decreasing (non-increasing), the limit of $(y_{n})$ must be the infimum of the set $\{y_{n}:n\in\mathbb{N}^{+} \}$. Thus,
$$\limsup_{n\to \infty}b_{n}=\lim_{n\to \infty}y_{n}=\inf\{y_{n}:n\in\mathbb{N}^{+} \}=y.$$
If $ y$ is the greatest lower bound of the set $\{y_{n}:n\in\mathbb{N}^{+} \}$, then any number greater than $y$ cannot be a lower bound for the set. If we let $ε$ be an arbitrary positive number, then $y+ε$ is not a lower bound for the set $$\{y_{n}:n\in\mathbb{N}^{+} \}.$$For $N$ sufficiently large ($N$ will depend on our choice of $ε$), $ y_{N}\ <y+ε$. We also know that for all $N$, $y_{N}\ge y$. $y_{n}$ is the supremum of the set $$\{b_{n},b_{n+1},b_{n+2},\cdots\}, $$so we can assume $y_{N}$ is the supremum of the set $$\{b_{N},b_{N+1},b_{N+2},\cdots\}.$$ If $n>N$, $b_{n}>y_{N}-ε$ (if there wasn't any terms of the set $\{b_{N},b_{N+1},b_{N+2},\cdots\}$ greater than $y_{N}-ε$ then $y_{N}-ε$ would be an upper bound for the set $$\{b_{N},b_{N+1},b_{N+2},\cdots\}$$ and we know this cannot be the case if $ε>0$ and $y_{N}$ is the least upper bound for the set $\{b_{N},b_{N+1},b_{N+2},\cdots\}$).
We now have the following:
$$y-ε\le y_{N}-ε<b_{n}<y_{N}<y+ε.$$
Hint:
Use the following statement (assume it has already been proven) : If for all $\epsilon>0 \ $the set $\{n\in\mathbb{N}^{+}:|x_{n}-x|<\epsilon \} \ $is infinite, then $\ (x_{n}) \ $ has a subsequence that converges to $\ x. \ $($\ (x_{n}) \ $is an infinite sequence of real numbers and $\ x\in\mathbb{R} \ $).
I don't really know where to go from here or whether all of the above is correct or not. Guidance would be greatly appreciated.
$$\lim y_n = y$$
That part actually looks like the answer to me, it's the definition of a convergent sequence.