If $A \subseteq X$ is open and $G \subseteq A$, then $G$ is open in $A$ iff $G$ is open in $X$

Question

I'm having some trouble solving this problem:

Show that if $A$ is an open set in $(X, d_X)$, then a subset $G$ of $A$ is open in $(A, d_A)$ if and only if it is open in $(X, d_X)$.

It seems that if $G$ is a subset of $A$, and $G$ is open in $A$, then $G$ would be open in any set $X$ for which $A$ is a subset of $X$. Clearly this is incorrect, hence the question, so if someone could straighten out my logic, it would be much appreciated.

Edit: For reference, this question is section 1.3, question 8C from this document: https://www.uio.no/studier/emner/matnat/math/MAT2400/v11/Metric.pdf

Answer 1

Assume $G$ is $A$-open.

Then $G=A\cap A'$ for some $X$-open $A'$.

Then $G$ is $X$-open since $A$ and $A'$ are both $X$-open.

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Assume $G$ is $X$-open.

Then $G$ is $A$-open since $G=A\cap G$ for $G$ $X$-open.