I'm trying to understand the proof of equivalent definitions of the limit superior.

Given the definition that $X=(x\_n)$ is a bounded sequence of real numbers and that the limit superior of $(x\_n)$ is the infimum of the set $V$ of $v\in \mathbb{R}$ such that $v<x\_n$ for at most a finite number $n\in\mathbb{N}$, I want to show the following implication, $(b)\implies (c)$, about a real number $x^\*$:

**$(b)$ If $\epsilon>0$, there are at most a finite number of $n\in\mathbb{N}$ such that $x^\*+\epsilon < x\_n$, but an infinite number of $n\in\mathbb{N}$ such that $x^\*-\epsilon<x\_n$.**

**$(c)$ If $u\_m=\sup\{x\_n:n\geq m\}$, then $x^\*=\inf\{u\_m:m\in\mathbb{N}\}=\lim(u\_m)$.**

The proof starts as follows:

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> Given $\epsilon>0$, $(b)$ implies for all sufficiently large $m$ that $u\_m<x^\*+\epsilon$. Therefore, $\inf\{u\_m:m\in\mathbb{N}\}\leq x^\*+\epsilon$.
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I'm trying to understand why this start of the proof is correct rigorously. Here's my attempt to derive that statement:

From part $(b)$, we know that for all $n$ greater than some $m\in\mathbb{N}$, $x\_n\leq x^\*+\epsilon$. Since $u\_m$ is the supremum of $\{x\_n:n\geq m\}$, for any given $\epsilon>0$, we have $u\_m-\epsilon<x\_n\leq u\_m$ for some $n\geq m$. I feel like these are the two pieces of data I need to combine into "$(b)$ implies for 'all sufficiently large $m$' that $u\_m<x^\*+\epsilon$" but I'm not sure how.

How does that imply anything about the set $\{u\_m:m\in\mathbb{N}\}$ and its infimum, in which $u\_m$'s index ranges over all natural numbers unlike in the previous "sufficiently large $m$" statement?