A relation $T$ is asymmetrical if and only if whenever $xTy$ holds, $yTx$ does not hold. That is, if and only if:
$$xTy\Longrightarrow y(\neg T) x.$$
So what you write makes no sense in terms of proving this. You write that you want to prove that if $xTy$ holds, then $x(\neg T)y$ is true; but this is not what you need to prove.
Also, your definition is that $xTy$ holds if and only if $x+1\leq y$; then you claim that $x(\neg T)y$ is equivalent to $y+1\leq x$; while that may be (see below) a true deduction, it's better to write what the condition means directly: if $xTy$ is equivalent to $x+1\leq y$, then $x(\neg T)y$ is equivalent to $\neg(x+1\leq y)$, which is equivalent to $x+1\not\leq y$.
Now, you don't specify what kind of numbers you are dealing with. For real numbers, $x+1\not\leq y$ is equivalent to $x+1\gt y$; but this is not the same as $y+1\leq x$ if you are dealing with real numbers, or with rational numbers. It's true for integers, but since you don't specify which kinds of $x$s you are dealing with, it could also be incorrect.
I think there are too many errors and unstated assumptions to make what you write a correct proof.
So, is this relation asymmetrical? For real numbers:
We want to prove that if $xTy$, then $y(\neg T)x$. So, assume that $xTy$. That means that $x+1\leq y$. We need to show that $y(\neg T)x$, meaning we need to show that $y+1\not\leq x$.
Indeed, we know that $x\lt x+1\leq y$; therefore, $x\lt y\lt y+1$. By transitivity, if $y+1\leq x$, then we would have $x\lt x$, which is impossible. Therefore, $y+1\not\leq x$. Therefore, $y(\neg T)x$.
Since the relation is asymmetrical for real numbers, it is also asymmetrical for rationals, integers, or natural numbers.