Disproving a inequality implication by contradiction. Let $x,y \in R$.
If $0 \leq y < x$ for all $x > 0$, then $y=0$.
Proof by contradiction: 
Assume the opposite that is;  "If $0 \leq y < x$ for all $x > 0$, then $y\neq0$".
         Subtract $x$ from each part of the inequality to get,
                      $0-x \leq y-x < 0$
         Then multiply through by -1 to get,
                      $x \geq y+ x > 0$
Since $x>0$, this implies a contradiction of the original statement, therefore we conclude that if $0 < y < x$ for all $x > 0$, then $y=0$.
Is my reasoning correct or is there something I can improve upon?
 A: If it's true for all $x>0$ that $0\leq y< x$ and if $y>0$ then take $x=y/2$ and you'll find a contradiction.
A: Lets take the contrarreciprocal $(\lnot q \Rightarrow \lnot p)$; 
If $y \neq 0$ then exist $x_0$ such that $y \geq x_0$. And you can always take $x_0 = y$.
A: This is not a direct answer to your question, but it may be helpful nonetheless: here are two direct proofs.  As in the question, all variables represent real numbers. $
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$

The first one is inspired by Levent's answer, here is a simple direct proof.
$$ \calc
\langle \forall x : x > 0 : 0 \le y < x \rangle
\op\equiv\hints{logic: extract $\;0 \le y\;$ which contains no $\;x\;$ out of $\;\forall x\;$}\hint{-- note that we silently use the fact that $\;\langle \exists x :: x > 0 \rangle\;$ here}
0 \le y \;\land\; \langle \forall x : x > 0 : y < x \rangle
\op\then\hint{choose $\;x := y\;$}
0 \le y \;\land\; (y > 0 \then y < y)
\op\equiv\hint{logic: simplify, using the fact that $\;y < y \;\equiv\; \false\;$}
0 \le y \;\land\; y \le 0
\op\equiv\hint{ordering: $\;\le\;$ is antisymmetric}
y = 0
\endcalc$$

Using the ordering property
$$\tag 0
\langle \forall x : a < x : b < x \rangle \;\equiv\; b \le a
$$
(which is basically transitivity in disguise) the proof becomes even simpler, and proves both directions at the same time:
$$ \calc
\langle \forall x : x > 0 : 0 \le y < x \rangle
\op\equiv\hint{logic: extract $\;0 \le y\;$ which contains no $\;x\;$ out of $\;\forall x\;$}
0 \le y \;\land\; \langle \forall x : x > 0 : y < x \rangle
\op\equiv\hint{property $\ref 0$, above, with $\;a,b := 0,y\;$}
0 \le y \;\land\; y \le 0
\op\equiv\hint{ordering: $\;\le\;$ is antisymmetric}
y = 0
\endcalc$$
