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: 
Let $f$ denote this power series and $(P\_n)$ denote the sequence of partial sums. Uniform convergence on $\Bbb R$ implies that $(P\_n)$ is uniformly Cauchy. I'll use this in two places. First, there must be some $N$ such that for all $n,m \ge N$, 

$$\|P\_n - P\_m\|\_{\infty} < 1$$

In particular, $\|P\_n - P\_N\|\_{\infty} < 1$ for all $n \ge N$, and since non-constant polynomials are unbounded, $P\_n - P\_N$ must be constant in $x$, say $:=b\_n$. Now,

$$|b\_n - b\_m| \le \|P\_n - P\_m\|\_{\infty} \to 0 \text{ as $n,m \to \infty$}$$

Where again I used that $(P\_n)$ is uniformly Cauchy. Hence $(b\_n)$ is Cauchy and so convergent, say to $b$. Thus, $f = b + P\_N$ is a polynomial.
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 $0<x<1$.