Proving by induction of $n$ that $\sum_{k=1}^n \frac {k+2}{k(k+1)2^{k+1}} = \frac{1}{2} \ - \frac1{(n+1)2^{n+1}} $

Your error is just after the sixth step from the bottom:

$$\frac{1}2 -\frac {1}{(n+1)2^{n+1}} +\frac{1}{(n+1)2^{n+1}} - \frac{1}{(n+2)2^{n+2}}=\frac{1}2 -\frac{1}{(n+2)2^{n+2}}$$

Then you are done.

You accidentally added the two middle terms instead of subtracting.

Using a telescoping sum, we get $$ \begin{align} \sum_{k=1}^n\frac{k+2}{k(k+1)2^{k+1}} &=\sum_{k=1}^n\left(\frac1{k2^k}-\frac1{(k+1)2^{k+1}}\right)\\ &=\sum_{k=1}^n\frac1{k2^k}-\sum_{k=2}^{n+1}\frac1{k2^k}\\ &=\frac12-\frac1{(n+1)2^{n+1}} \end{align} $$