To prove $\sum_{n=0}^\infty \binom{r}{x}\binom{N-r}{n-x}=\binom{N}{n}.$ To prove $$\sum_{x=0}^n \binom{r}{x}\cdot \binom{N-r}{n-x}=\binom{N}{n}.$$
I tried comparing the coefficients of
$(1+x)^{(n+k)} = (1+x)^n(1+x)^k$
but couldn't reach the answer. 
 A: (replacing the $x$s in your summation with $k$s to facilitate generating function methods)
Fix $N$, and let $f(n)=\sum_k\binom{r}{k}\binom{N-r}{n-k}, F(x)=\sum_nf(n)x^n$. Then:
$$F(x)=\sum_{n,k}\binom{r}{k}\binom{N-r}{n-k}x^n=\sum_{n,k}\binom{r}{k}x^k\binom{N-r}{n-k}x^{n-k}$$
Let $m=n-k$, so that
$$F(x)=\sum_{m,k}\binom{r}{k}x^k\binom{N-r}{m}x^{m}=\left(\sum_k\binom{r}{k}x^k\right)\cdot\left(\sum_m\binom{N-r}{m}x^m\right)$$
Spotting these as binomial series, we have $F(x)=(1+x)^r(1+x)^{N-r}=(1+x)^N$, which allows us to recover $f(n)=\binom{N}{n}$.
Alternatively, let's pick a subset of size $n$ from ${1,2,...,N}$, and of the $n$ we pick, we take $k$ of them from ${1,2,...,r}$ and the remaining $(n-k)$ from ${r+1,...,N}$. Now sum over all possible $k$.
A: Hint your efforts seem to be algebraic thats why this hint see Vandermonde's identity on wikipedia which states $$\sum {r\choose x}{N-r\choose n-x}={N\choose n}$$
A: I suspect you wish to show the following.
$$\sum\limits_{x=0}^n \dbinom{n}{x}\dbinom{N-n}{n-x} = \dbinom{N}{n}$$
The right-hand side counts distinct ways to select $n$ objects from a list of $N$ distinct objects.
The left hand side counts the ways to select $x$ of the first $n$ objects in the list and $n-x$ of the remaining $N-n$, and to do so for $x$ from $0$ to $n$.
These are merely different approaches to counting the same task, so their counts must be equal.
Which was to be show. 
$\Box$
A: 
We  obtain for $0\leq   r\leq N$
  \begin{align*}
(1+x)^{N}&=(1+x)^r(1+x)^{N-r}\\
&=\sum_{k=0}^r\binom{r}{k}x^k\sum_{l=0}^{N-r}\binom{N-r}{l}x^l\\
&=\sum_{n=0}^{N}\left(\sum_{{k+l=n}\atop{k,l\geq 0}}\binom{r}{k}\binom{N-r}{l}\right)x^n\\
&=\sum_{n=0}^{N}\left(\sum_{k=0}^n\binom{r}{k}\binom{N-r}{n-k}\right)x^n\\
\end{align*}
  Since
  \begin{align*}
(1+x)^{N}=\sum_{n=0}^{N}\binom{N}{n}x^n
\end{align*}
  we compare coefficients and conclude
  \begin{align*}
\sum_{k=0}^n\binom{r}{k}\binom{N-r}{n-k}=\binom{N}{n}
\end{align*}

