Assume that a sequence $\{x_i\}_{i=0}^\infty$ of Newton's method $x_{i+1}=x_i-\frac{f(x_i)}{f'(x_i)}$ is convergent and set $x_\infty = \lim_{i\to\infty} x_i$. Furthermore assume that $f'(x_\infty) \neq 0$. How is it possible to show that $x_\infty$ is an solution to $f(x)=0$?
My solution for this were assuming that $x_{\infty+1}$=$x_\infty$ so $$x_{\infty+1}=x_\infty-\frac{f(x_\infty)}{f'(x_\infty)}$$ $$0=-\frac{f(x_\infty)}{f'(x_\infty)}$$ $$0=-f(x_\infty)$$ $$f(x_\infty)=0$$
The following proof works if we assume that $\{x_i\}_{i\geq 0}$ is a convergent sequence such that $x_{i+1}=x_i-\frac{f(x_i)}{f'(x_i)}$ for all $i\geq 0$ (this assumption wasn't on your initial post, hence my comment) and $f$ and $f'$ are both continuous in some interval containing $x_\infty$ (i.e. $f$ is continuously differentiable in $x_\infty$).
Your intuition that "$x_{\infty}=x_{\infty+1}$" is good, but doesn't yet constitute a proof. We can use the fact that $\displaystyle \lim_{i\to\infty} x_i=\lim_{i\to\infty} x_{i+1}$ (see here; this stems from the fact that $\{x_{i+1}\}_{i\geq 0}$ is a subsequence of $\{x_i\}_{i\geq 0}$) to deduce the following: $$\lim_{i\to \infty} x_{i+1}=\lim_{i\to \infty}\left(x_i-\frac{f(x_{i})}{f'(x_i)}\right) ~~\implies~~x_{\infty}=x_{\infty}-\frac{f(x_{\infty})}{f'(x_{\infty})},$$ by sum and fraction of limits and the convergence of $\{f(x_i)\}_{i\geq 0}$ and $\{f'(x_i)\}_{i\geq 0}$ if $f,f'$ are continuous and $f'(x_{\infty})\neq 0$. Substracting $x_{\infty}$ on both sides and multiplying by $-f'(x_{\infty})$, we obtain: $$-\frac{f(x_{\infty})}{f'(x_{\infty})}=0 ~~~~\text{and thus}~~~~ f(x_{\infty})=0.$$ That is, $x_{\infty}$ is a solution for $x$ in $f(x)=0$.
I'd like to remark (as in my comment) that the Newton-Rhapson method does not always converge to a root; this proof required fairly strong assumptions. Here's an interesting discussion about this.
