Question: Let $\sum\_{n=0}^\infty a\_n \left( x-x\_0 \right)^n$ be a power series that converges uniformly over all $x\in \mathbb R$. Prove there exists $N\in\mathbb N$ such that for all $n>N,\; a\_n = 0$.

I fail to see how this is true.
From the radius of convergence formula 
$$\frac{1}{R}=0=\limsup\_{n\rightarrow\infty}\sqrt[n]{\left|a\_n\right|}$$
Now for all $n$ choose $a\_n=\frac{1}{n^n}\neq0$ and we get $a\_n>a\_{n+1}$ so that
$$\limsup\_{n\rightarrow\infty}\sqrt[n]{\left|a\_n\right|}=\lim\_{n\rightarrow\infty}\frac{1}{n}=0$$ We therefore got a convergent series over all the reals, thus uniformly convergent in every $[a,b]\in \mathbb R$ contradicting the above statement...

Answer: 
Uniform convergence of the power series of $f$ is equivalent to the fact that the sequence 
$$
f\_n(x)=\sum\_{k=0}^na\_kx^k, \quad n\in\mathbb N,
$$
is uniformly convergent and equivalently uniformly Cauchy in $\mathbb R$, i.e., for every $\varepsilon>0$, there is an $N$, such that, $$m,n\ge N\quad\Longrightarrow\quad
\sup\_{x\in\mathbb R}\lvert\, f\_m(x)-f\_n(x)\rvert<\varepsilon.\tag{1}$$

Assume that the power series contains infinitely many terms; say $a\_{k\_n}\ne 0$, $n\in\mathbb N$, let $\varepsilon>0$ and $N$, such that $(1)$ holds. Then, for $k\_n\ge N$ we have
$$
\sup\_{x\in\mathbb R}\lvert\, f\_{k\_n}(x)-f\_{k\_n-1}(x)\rvert=\sup\_{x\in\mathbb R}\lvert a\_{k\_n}x^{k\_n}\rvert=\infty.
$$
Contradiction. Hence, the power series contains only finitely many terms.
It appears that you're under the impression that "uniformly convergent" means convergent everywhere in $\mathbb R$.

Consider the sequence of functions $x\mapsto x^n$ on the interval $0<x<1$.

For every value of $x$ in the interval you have $x^n\to0$ as $n\to\infty$. That is what it means to say this sequence of functions converges **pointwise** on the interval $0<x<1$.

But the uniform distance from $x\mapsto x^n$ to $x\mapsto0$ is $\displaystyle\sup\_{0<x<1} |x^n - 0| = 1$, and that does not approach $0$ as $n\to\infty$. That is what it means to say this sequence of functions does **not** converge **uniformly** on the interval $0<x<1$.