Showing if $\limsup x_n = 1$ then $x_n < 1.01$ eventually 
If $\limsup x_n = 1$ then $x_n < 1.01$ for sufficiently large $n$

I was given this question. I believe that it is true, since eventually the sequence is close to $1$ but it cannot be greater than its $\limsup$. Here is my attempt at a proof, let me know if I've gone astray.
For $\varepsilon = 0.01$ there exists an $N$ so that $\forall n \geq N$:
$|\sup \{x_m : m \geq n\} - 1| < 0.01$. 
In particular, $|\sup \{x_m : m \geq N\} - 1| < 0.01$, then 
$\sup \{x_m : m \geq N\} < 1.01$ and since the $\sup \{x_m : m \geq N\}$ is the least upper bound for all $n \geq N$ then 
$x_n \leq \sup \{x_m : m \geq N\} < 1.01$, for $n \geq N$.
 A: $\limsup_n x_n$ is normally defined as $\lim_n \sup\{x_m: m \ge n\}$, which exists because (setting $s_n = \sup \{x_m: m \ge n\}$) $s_n$ is a decreasing sequence in $n$ (a sup of a smaller set can only be smaller, so $s_{m+1} \le s_m$ for all $m$, and all decreasing sequences have a limit in $[-\infty,+\infty]$, the extended reals. 
Using that definition, your proof of the fact that eventually all terms are $< 1.01$ (or $1+\varepsilon$ for any $\varepsilon$) works and is quite correct.
But we can also see the $\limsup_n x_n$ as the maximal value that a limit of a subsequence of $(x_n)$ can have and likewise $\liminf_n x_n$ as the smallest subsequential limit.
Then if the assertion about the $x_n$ being eventually smaller than $1+\varepsilon$ would fail, there would be for every $n_1$ some $n_2 > n_1$ with $x_{n_2} \ge 1+\varepsilon$, and this would define a subsequence (maybe using compactness of $[1+\varepsilon, +\infty]$ if necessary) of $x_n$ with a subsequential limit $L \ge 1+\varepsilon$ as well, which cannot be as $\limsup_n x_n = 1$ is the largest subsequential limit. So we can only have finitely many terms $x_n \ge 1+\varepsilon$.
