Is my reasoning correct here regarding uniform convergence? Let $f_n(x) = x^n$.
Is this sequence of functions uniformly convergent on the closed interval $[0, 1]$?
My reasoning
Consider $0 \le x < 1$. $f_n(x)$ converges pointwise to the zero function on this interval.
$f_n(x)$ is uniformly convergent if there exists some $N$ such that for $n \ge N$
$$|f_n(x) - f(x)| < \epsilon$$ for all $x$, $\epsilon > 0$.
In particular, $|f_N(x) - f(x)| < \epsilon$.
Choose $\epsilon = \frac{1}{4}$
So is $|f_N(x) - f(x)| < \epsilon$ for every $x$?
I.e. is $|f_N(x)| = |x^N| = x^N < \frac{1}{4}$ for every $x$?
Well I claim that no matter what $N$ we take, if we take $$x \ge 4^{\frac{-1}{N}}$$
Then $x^N \nless \frac{1}{4}$ so the convergence isn't uniform.
Is that reasoning correct?

You're are certainly very close to a full (and correct!) solution. Let's first straighten a couple things out: 
1) The $N$ you choose can depend on $\epsilon$. That is, uniform convergence of $f_n \to f$ means: given an $\epsilon >0$ there exists an $N \in \mathbb{N}$ such that for all $x$ in the domain of $f$ it holds that  $|f_n(x) - f(x)| < \epsilon$. 
2) Let's take your value of $\epsilon = 1/4$. Then, what you can say is that there exists an $N$ (depending on the value $1/4$) such that for every $n \geq N$ and any $x \in [0,1]$, $|f_n(x) -f(x)|<\epsilon$ (if it does converge uniformly). However, as you noted if $x \in (4^{-1/n},1)$ then $|f_n(x) - 0| \geq (4^{-1/n})^n = 1/4 = \epsilon$. This proves that $f_n$ does not converge uniformly to the function $f(x)$ where $f(x) \equiv 0$ on $[0,1)$.
3) Since if $f_n \to f$ uniformly and $f_n \to g$ pointwise, it must be that $f = g$. As you said, $f_n \to 0$ pointwise on $[0,1)$ so the only candidate function $f(x)$ is the function $f(x) \equiv 0$ on $[0,1)$. 
Now your proof is done because you have found a contradiction! 
There is another way to prove this as well. Here's how: Note that each $f_n$ is continuous on $[0,1]$. Therefore, if $f_n \to f$ uniformly, it must be that $f$ is continuous on $[0,1]$. However, pointwise $f_n$ converges to the function $$f(x) = \begin{cases} 0 & x \in [0,1) \\ 1 & x=1 \end{cases}$$ which is not continuous. Since uniform convergence implies pointwise convergence and $f_n \to f$ pointwise with $f$ not continuous, it must be that $f_n$ does not converge uniformly.
