Prove For all integers $x$ if for all natural numbers $y$, $x$ does not divide $y$, then $x = 0$.
I start by saying that $x\neq 0$ then $x\mid y$ there is exists an integer $d$ such that $xd=y$ if $x$ is not zero and $d$ is not $0$ then $x\mid y$ for all integers $x$ and natural numbers $y$.
Does this proof make sense and if not how can I prove my statement? THanks!
The contrapositive is: if $x\neq0$ then $x|y$ for some natural number $y$. You are attempting to prove: if $x\neq0$ then $x|y$ for all natural numbers $y$. That is false, and of course your argument doesn't prove it. A possible correction is below.
Let $x\neq0$ be any integer, and set $y:=x$ if $x>0$ and $y:=-x$ if $x<0$. Then $y$ is a natural number and $x|y$. By contrapositive, if $x$ does not divide any natural number then $x=0$.