Proving $\lim\sup x_n = \lim\inf x_n$ for $x_n$ Cauchy, alternate 
Suppose $x_n$ is a Cauchy sequence. Prove that the $\lim\sup x_n = \lim\inf x_n$. 

I made some serious mistakes when first approaching this, wanted some feedback on the proof to see if there were any errors, that is why I posted another question. Again, cannot use the fact that all Cauchy converge since we haven't shown this result yet.
Define $b_n = \sup\{x_k : k \geq n\}$ and $a_n = \inf\{x_k: k \geq n\}$. Finally, denote $c_n = \sup\{x_i - x_j: i,j \geq n\}$. I've recently proved in another question that $c_n = b_n - a_n$. Note that $c_n$ is a decreasing sequence and $c_n \geq 0$ for all $n$
Suppose $\varepsilon > 0$, since $x_n$ is Cauchy then there is an integer $N$ so that $|x_i -x_j| < \frac{\varepsilon}{2}$ for $i,j \geq N$.  Thus,
$b_N - a_N = c_N \leq \frac{\varepsilon}{2} < \varepsilon$. Since, $c_n$ is decreasing and bounded below by $0$, we have:
$-\varepsilon < 0 \leq c_n < \varepsilon$ for all $n \geq N$ and thus $|c_n| < \varepsilon$ .
Therefore, $c_n \rightarrow 0$ as $n \rightarrow \infty$ and so $c_\infty = 0$. Since, $c_\infty = b_\infty - a_\infty$, then $b_\infty = \lim\sup x_n = \lim\inf x_n = a_\infty$.
 A: Basically your proof is correct, but maybe you should clarify some steps and consider different wording on other.
That $c_N = b_N-a_N$ might need some explanation.
That $c_n$ decreases might sound obvious, but also there one may note that the $\sup$ over a subset is no larger than $\sup$ over the full set (to nitpick one might either have to know that the sets are non-empty or that $\sup\emptyset=-\infty$).
You don't have to write $-\epsilon < 0$ in the penultimate paragraph.
The last paragraph should probably be reworded. The symbols $a_\infty$, $b_\infty$ and $c_\infty$ sounds like very sloppy reasoning. We know that $c_n = b_n-a_n$ and $\lim c_n = 0$ so $\lim (b_n-a_n) = 0$.
One may need to complete the last with the fact that by definition $\limsup x_n = \lim b_n$, and that it exists (because cauchy sequences are bounded(*) and therefore $b_n$ is and it's monotone) and the corresponding for $\liminf$ this means that $0 = \lim(b_n-a_n) =  \lim b_n - \lim a_n$.
(*) Cauchy sequences are bounded because take $\epsilon=1$ and you have an $N$ such that especially $|x_n-x_N| < \epsilon = 1$ for all $n>N$. So for $n>N$ we have $|x_n| < |X_N+1|$ and for $n\le N$ we have $|x_n| < \max(|x_1|, |x_2|, \cdots, |x_N|)$.
