(1) Proof: Since $a_n$ is the arithmetic mean of $n$ and $S_n$, we have
$$2a_n = n + S_n \tag{1}$$
When $n=1$, $a_1 = S_1 = 2a_1 - 1$, which gives $a_1 = 1$.
When $n \geq 2$, we have
$$2a_{n-1} = (n-1) + S_{n-1} \tag{2}$$
Subtracting equation (2) from equation (1), we get
$$2a_n - 2a_{n-1} = 1 + a_n$$
which simplifies to
$$a_n = 2a_{n-1} + 1$$
Then,
$$a_n + 1 = 2(a_{n-1} + 1)$$
Thus, ${a_n + 1}$ is a geometric sequence with the first term $a_1 + 1 = 2$ and common ratio $r = 2$.

(2) Now let's find the sum $T_n$:
$${ \frac {1}{log_{2}(a_{n}+1)log_{2}(a_{n+2}+1)}} = \frac {1}{log_{2}2^{n} \cdot log_{2}2^{n+2}} = \frac{1}{n(n+2)}$$
Using partial fractions, we can rewrite $\frac{1}{n(n+2)}$ as:
$$\frac{1}{n(n+2)} = \frac{1}{2}\left(\frac{1}{n} - \frac{1}{n+2}\right)$$

Now let's find the sum of the first $n$ terms, $T_n$:
\begin{align*}
T_n &= \frac{1}{2}\left[ \left(1 - \frac{1}{3}\right) + \left(\frac{1}{2} - \frac{1}{4}\right) + \left(\frac{1}{3} - \frac{1}{5}\right) + \cdots + \left(\frac{1}{n-1} - \frac{1}{n+1}\right) + \left(\frac{1}{n} - \frac{1}{n+2}\right) \right] \\
&= \frac{1}{2}\left[ 1 + \frac{1}{2} - \frac{1}{n+1} - \frac{1}{n+2} \right] \\
&= \boxed{\frac{3}{4} - \frac{1}{2}\left(\frac{1}{n+1} + \frac{1}{n+2}\right)}
\end{align*}