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: 
Suppose that a sequence $f\_n(x)$ converges pointwise to the function $f(x)$ for all $x\in S$.

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> **NEGATION OF UNIFORM CONVERGENCE**
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> The sequence $f\_n(x)$ fails to converge uniformly to $f(x)$ for $x\in S$ if there exists a number $\epsilon>0$ such that for all $N$, there exists an $n\_0>N$ and a number $x\in S$ such that $|f\_{n\_0}(x)-f(x)|\ge \epsilon$.
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Now, let $f\_n(x)=a\_nx^n$ with $\lim\_{n\to \infty}f\_n(x)=0$ for all $x\in \mathbb{R}$. Certainly, either $a\_n=0$ for all $n$ sufficiently large or for any number $N$ there exists a number $n\_0>N$ such that $a\_{n\_0}\ne 0$.

Suppose that the latter case holds. Now, taking $\epsilon=1$, we find that

$$|a\_{n\_0}x^{n\_0}|\ge \epsilon$$

whenever $|x|\ge |a\_{n\_0}|^{-1/n\_0}$. And this negates the uniform convergence of $f\_n(x)$.

And inasmuch as the sequence $f\_n(x)$ fails to uniformly converge to zero, then the series $\sum\_{n=0}^\infty f\_n(x)$ fails to uniformly converge.

Note for the example for which $a\_n=n^n$ we can take $x>1/n$.

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**NOTE:**

If $a\_n=0$ for all $n>N+1$, then we have

$$\sum\_{n=0}^\infty a\_nx^n = \sum\_{n=0}^N a\_nx^n$$

which is a finite sum and there is no issue regarding convergence.
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 $0<x<1$.