(I) To prove: By using the Cauchy-Schwarz inequality, we have $(a+b+c)^{2} \leq (1^{2}+1^{2}+1^{2})(a^{2}+b^{2}+c^{2})$, which simplifies to $(a+b+c)^{2} \leq 3$, and hence $|a+b+c| \leq \sqrt{3}$. $\boxed{|a+b+c| \leq \sqrt{3}}$

(II) For the solution: The inequality $|x-1|+|x+1| \geq (a+b+c)^{2}$ is always satisfied for all real numbers $a$, $b$, $c$. From part (I), we know that $|x-1|+|x+1| \geq 3$. We analyze this inequality in three cases depending on the value of $x$:

- Case 1: If $x \geq 1$, both $|x-1|$ and $|x+1|$ are equal to $x-1$ and $x+1$, respectively, so the inequality becomes $2x \geq 3$, which gives us $x \geq \frac{3}{2}$.
- Case 2: If $x \leq -1$, both $|x-1|$ and $|x+1|$ are equal to $-(x-1)$ and $-(x+1)$, respectively, so the inequality becomes $-2x \geq 3$, leading to $x \leq -\frac{3}{2}$.
- Case 3: If $-1 < x < 1$, $|x-1| = -(x-1)$ and $|x+1| = x+1$, combining these gives $2 \geq 3$, which is not possible.

Combining the results of the valid cases, we find that $x$ is either $\geq \frac{3}{2}$ or $\leq -\frac{3}{2}$. Therefore, the range of the real number $x$ is $(-\infty, -\frac{3}{2}] \cup [\frac{3}{2}, +\infty)$. $\boxed{(-\infty, -\frac{3}{2}] \cup [\frac{3}{2}, +\infty)}$