Proving $f_n(x) = x^n$ for $x \in [0,1]$ is not a uniformly Cauchy sequence I know this can be argued more succinctly by citing theorems on uniform convergence.
I have also seen answers to this question on this page, but I made a somewhat different (I think) attempt of my own.
I wanted to take a stab at proving this using only the negated definition of a uniformly Cauchy sequence, namely

There is a real $\varepsilon >0$ such that for each $N \in \mathbb{N}$ there is $n \geq N$ such that for some $x \in [0,1]$ the following holds: $|f_N (x) - f_n(x)| \geq \varepsilon $.

Below is my attempt, I would be much interested in feedback on whether it can be improved or if I have committed any logical errors. Thanks!

For $x \in \{0,1\}$ the difference $|f_N(x) - f_n(x)|$ is zero, so henceforth we will consider $x \in (0,1)$.
Take some positive real $\varepsilon < 1$. Let $N$ be given, upon which we have the positive real  $\varepsilon^{1/N} < 1$.
Let $n \geq N$ and write $n = N + k$ where $k \geq 0$.
For $x \in (0,1)$, which soon will be chosen judiciously, we have

$|f_N (x) - f_n (x) | = |x^N - x^N x^k| = x^N (1-x^k)  $

$x^k$ can be made arbitrarily small, so take $k$ large enough such that $x^k < 1/2$. Then we have $1-x^k > 1/2$.
Thus

$|f_N (x) - f_n (x) | = x^N (1-x^k) > x^N \frac 12  $

Take now $ x = \varepsilon^{1/N}$. Then

$|f_N (x) - f_n (x) | > x^N \frac 12 = \varepsilon /2 $

Hence we have demonstrated  a positive real $\varepsilon/2 < 1/2$ such that  for each natural $N$ there is an $n \geq N$ such that for $x = \varepsilon^{1/N} \in (0,1)$ we have $|f_N (x) - f_n(x)| > \varepsilon/2$. $\square$
 A: I think you "switched" the logical order of the argument you need to make in the last part of the proof.
My guess is that your argument fails here:
...$x^k$can be made arbitrarily small, so take $k$ large enough such that $x_k<1/2$. Then we have $1−x_k>1/2$....
What do you mean with it?For which $x$ is valid that inequality? Note that you haven't fixed $x$ at that moment of the proof so you will need that inequality to be valid for wherever value you want to give to $x$...
Here below are some confused hints to what I suppose is a solution:
....For x∈(0,1), which soon will be chosen judiciously, we have
$$|f_N(x)−f_n(x)|=|x^N−x^Nx^k|=x^N(1−x^k)$$...
What we need to do now is to choose(fix) $x$ in such a way that $x^N=\epsilon +\delta$ with $\delta$ some positive number  $\delta\gt0$, so now $x$ is fixed, and we can still "pump up" $k$ arbitrary high... How high we need $k$? Well, we need it enough high so that the term $1-x^k$ (note that here we have already fixed $x$ so we will consider a number that depends on $\epsilon$), is enough "near" 1 so that $x^N(1−x^k)\gt \epsilon$
...
At the moment I'm typing on my phone, so excuse me any errors that you will find and the horrific editing.
