$\lim x_n^{x_n}=4$ prove that $\lim x_n=2$ [duplicate]

There are interesting elements in your proof. However, the way you state it can be improved:

1. You mention Cauchy criteria, but what you use is not Cauchy criteria.
2. You suppose that $\lim x_n$ is not equal to $2$. You cannot say $|x_n-2|>\epsilon$ without mentionning of what $n$ your're speaking. You should state that you have a subsequence $(x_{\beta_n})$ for which $|x_{\beta_n}-2|>\epsilon$.
3. From there you have a subsubsequence with either $x_{\beta_n^\prime}>\epsilon+2$ or $x_{\beta_n^\prime}<-\epsilon+2$.
4. Then you can follow with your other arguments.

However, I think that a totally different proof can use the map $f : x \mapsto x^x$, that is continuous, strictly increasing for $x \in (1,+\infty)$ and therefore invertible around $x=2$ with a continuous inverse. The conclusion is then straightforward.