Proof:
(I) From $a_{n+1}^2 + 2a_{n+1} = a_n + 2$ (1), we get $a_{n+2}^2 + 2a_{n+2} = a_{n+1} + 2$ (2).

Subtracting equation (1) from equation (2), we obtain $(a_{n+2}^2 - a_{n+1}^2) + 2(a_{n+2} - a_{n+1}) = a_{n+1} - a_n$, which simplifies to $(a_{n+2} - a_{n+1})(a_{n+2} + a_{n+1} + 2) = a_{n+1} - a_n$.

Since $a_n > 0$, $(a_{n+2} - a_{n+1})$ and $(a_{n+1} - a_n)$ have the same sign.

As $(a_2 + 1)^2 = a_1 + 3 = 5$, we have $a_2 = \sqrt{5} - 1$, and $(a_2 - a_1) = \sqrt{5} - 3 0$, we have $a_n - 1 > 0$, which means $a_n > 1$.

As a result, $(a_{n+1} + 3) > 4$, and $\frac{1}{a_{n+1} + 3} < \frac{1}{4}$. Thus, $|a_{n+1} - 1| < \frac{1}{4} |a_n - 1|$.

(III) From (II), we know that when $n \geqslant 2$, $|a_n - 1| = |a_1 - 1| \times \frac{|a_2 - 1|}{|a_1 - 1|} \times \frac{|a_3 - 1|}{|a_2 - 1|} \times \cdots \times \frac{|a_n - 1|}{|a_{n-1} - 1|} < \frac{1}{4^{n-1}} |a_1 - 1| = \frac{1}{4^{n-1}}$.

Since $a_1 - 1 = 1$, we have $|a_n - 1| \leqslant \frac{1}{4^{n-1}}$, $n \in \mathbb{N}^*$.

Therefore, $|a_1 - 1| + |a_2 - 1| + |a_3 - 1| + \cdots + |a_n - 1| \leqslant 1 + \frac{1}{4} + \frac{1}{4^2} + \cdots + \frac{1}{4^{n-1}} \leqslant \frac{1 - \frac{1}{4^n}}{1 - \frac{1}{4}} = \frac{4}{3} \left( 1 - \frac{1}{4^n} \right) < \boxed{\frac{4}{3}}$.