There is already a solution to this question on the site, but I just wanted to verify that my thought process to arrive at my solution was correct as well.
The question asked to show that the sequence of functions $f_{n}(x) = x^{n}$ does not converge uniformly to the function:
$$f(x) = \begin{cases} 0 &,\ 0 \leq x < 1 \\ 1 &,\ x = 1 \end{cases} $$
To work this out I approached things by using the case where $x \in [0,1)$. And as such I am left to find a scenario that staisfies:
$$|f_{n}(x) - f(x)| = |x^{n} - 0| \geq \epsilon$$
After algebra I construct an $x$ that would satisfy my desired result:
$$x \geq \sqrt[n]{\epsilon}$$.
So this means that as long as I choose any point $x$ in the interval I would arrive at this conclusion. Now this is satisfied as long as $0 < \epsilon < 1$. If $\epsilon > 1$ then choose my $x$ such that it satisfies $x \geq \sqrt[n]{\epsilon - \lfloor \epsilon \rfloor}$ (there is definitely an easier choice but I forgot what it was and I'm curious to see it).
So to now prove the statement, I would begin by choosing an $\epsilon$. For this let's say that $\epsilon = \frac{1}{2}$. Letting $x$ be such that it satisfies the property: $x \geq \sqrt[n]{\epsilon}$, this then means:
$$|f_{n}(x) - f(x)| = \bigg|\bigg(\bigg(\frac{1}{2}\bigg)^{\frac{1}{n}}\bigg)^{n} - 0 \bigg| = \frac{1}{2} \geq \epsilon = \frac{1}{2}$$
Which would satisfy the negation of the definition of uniform convergence.
Is this the way to approach such a question? Also how to treat the case of $\epsilon > 1$ more cleanly?
Since you are going to apply your approach to $\varepsilon=\frac12$, there is no need for you to say what you would do if $\varepsilon>1$. And, if you felt that need, then you would also have to say what you would do if $\varepsilon=1$.
Other than that (and the fact that apparently you use the letters $n$ and $N$ for the same thing) it is correct.