Question: .

my steps assume $d=2$ and $n=1$, then $2$ divides $1 = 1/2$
 if d does not divide $(n +1)$ then $2/1+1 = 1$
 $1/2$ is not equal to $1$
 hence it is false!

Answer: 
Let $d$ divide $n$.

Therefore, for some natural $c$, $n=cd$.

Assume $d$ also divides $n+1$, and therefore, for some natural $f$, $n+1=fd$.

Since $n+1>n$ it follows $f>c$ so $f=c+g$ for some natural $g$.

From this it follows $n+1=d(c+g)=cd+gd$. 

However $cd=n$ so we have $n+1=n+gd \Rightarrow 1=gd$.

This is a contradiction! We are given $d>1$ and we have defined $g$ as a natural number, so $gd>1$, which contradicts our result $gd=1$. 

Thus, for all $d>1$ and natural $n$ if $d$ divides $n$ then $d$ does not divide $n+1$.
If $d$ divides $n+1$ as well

$d$ must divide $1\cdot(n+1)+(-1)\cdot n$