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> Consider $A=\{q \in \mathbb Q:q^2 \geq2\}$ and the metric $d(x,y)=|x-y|$. 
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Then considered as a subset of the metric space $(\mathbb R,d) A$ is not open since none of the points of $A$ is an interior point of A since every nbd in $\mathbb R$ contains irrational numbers not belonging to $A$. Also $\sqrt2 $ is a limit point of $A$ not contained in $A$ so $A$ is not closed either.

However, when considered as a subset of the metric space $(\mathbb Q,d)$ it is both open and closed. Since now the nbds only contain rationals, we can find one of every point that lies inside $A$. And now since the parent space is $\mathbb Q$, $\sqrt 2$ is no longer considered a limit point and the remaining limit points are all the members of $A$ thus lying inside $A$ and making it closed.

Is this reasoning correct? Am I right in assuming that a subset is open or closed depending on the set it is considered a subset of?