Is this proof false?

Could please someone help me? I proved that

$X \subseteq [0,1]$ has Lebesgue outer measure $\mu^\ast (X)> 0$ iff $\forall Y$ with $\mu (Y) = 1$, $X \cap Y \neq \varnothing$

But both directions of proof seem to be the same. This is what I wrote:

Let $\mu^\ast (X) > 0$. If there was $Y \subseteq [0,1]$ with $\mu (Y) = 1$ and $X \cap Y = \varnothing$ then $X \subseteq [0,1] \setminus Y$ and then $\mu^\ast (X) \le \mu^\ast ([0,1] \setminus Y) = 0$ which is contradiction.

Other direction: Assume there is $Y$ such that $Y \cap X = \varnothing$ then $X \subseteq [0,1] \setminus Y$ and then $\mu^\ast (X) \le \mu^\ast ([0,1] \setminus Y) = 0$.

Is this valid proof? Or did I prove same direction twice? Thank you for correcting me.

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$$X\subseteq[0,1]\textrm{ and }\mu^*(X)>0\implies X\cap Y\neq\emptyset \textrm{ for all measurable }Y\subseteq [0,1]\textrm{ with }\mu(Y)=1.$$
You have shown it directly the first time and the contrapositive the second time. For the other direction, let $X\subseteq[0,1]$ and $\mu^*(X)=0$ and find a measurable set $Y\subseteq[0,1]$ with $\mu(Y)=1$ such that $X\cap Y=\emptyset$.