Proof:
(1) From $a_{n+1}-2a_n=2^n$,
dividing both sides by $2^{n+1}$, we get
$\frac{a_{n+1}}{2^{n+1}} - \frac{a_n}{2^n} = \frac{1}{2}$,
which shows that the sequence $\left\{\frac{a_n}{2^n}\right\}$ is an arithmetic sequence with the first term $\frac{1}{2}$ and common difference $\frac{1}{2}$;
thus, $\frac{a_n}{2^n} = \frac{1}{2} + \frac{1}{2}(n-1) = \frac{n}{2}$,
hence, $a_n = n\cdot 2^{n-1}$;
(2) $b_n = \frac{(n+2)2^{n-1}}{a_na_{n+1}} = \frac{(n+1)\cdot 2^n - n\cdot 2^{n-1}}{a_na_{n+1}}$
$= \frac{a_{n+1}-a_n}{a_na_{n+1}} = \frac{1}{a_n} - \frac{1}{a_{n+1}}$,
thus, $S_n = b_1 + b_2 + \ldots + b_n$
$= \frac{1}{a_1} - \frac{1}{a_2} + \frac{1}{a_2} - \frac{1}{a_3} + \ldots + \frac{1}{a_n} - \frac{1}{a_{n+1}}$
$= \frac{1}{a_1} - \frac{1}{a_{n+1}} = 1 - \frac{1}{(n+1)\cdot 2^n} < 1$.

Thus, the final answers are:
(1) The general formula for $\{a_n\}$ is $\boxed{a_n = n\cdot 2^{n-1}}$.
(2) It is proven that $\boxed{S_n < 1}$.