The underlying set of an induced metric space is both closed and open in itself

Question

I have a following reasoning in metric space:

Let $(X,d)$ be a metric space and $A \subseteq X$. Let $(A, d')$ be the metric space induced from $d$. Then $A = X \cap A$. Because $X$ is both closed and open in $X$, $A$ is both closed and open in $A$.

Could you please confirm if it is correct? Thank you so much!

Answer 1

Logically sound, but there's something that bothers me. You seem happy to assume that, for any metric space $X$, $X$ is closed and open within itself. You also seem happy to call the subset $A$ of $X$ a metric space when equipped with the induced metric. This being the case, surely proving $A$ is closed and open in $A$ is redundant, when you're assuming this is true for an arbitrary metric space $X$ to begin with?

Answer 2

Yes. The underlying set for any topological space is both closed and open in itself. It will still be so for metric spaces, and it will still be so for subspaces.