Let $(X,d)$ be a metric space and $A\subset X$ a subset equipped with the induced metric $d_{A}$. Then the open subsets of $(A,d_{A})$ are exactly the intersections of open subsets of of $(X,d)$ with $A$: $B \subset A$ is open in $(A,d_{A})$ iff there exists an open subset $Y \subset X$, so that $B = A \cap Y$.
My proof of "$\implies$" is pretty straight forward:
Let $B\subset A$ be an open subset. Define \begin{equation*} U_{\varepsilon}^{Z}(x) := \{ y \in Z: d_A(x,y) = d(x,y) < \varepsilon\} \end{equation*} for any subset $Z \in X$. Because $B$ is open, for every $x \in B$ there exists an $\varepsilon_{x} > 0$ so that \begin{equation*} B\supset U^{A}_{\varepsilon_{x}}(x) = A\cap U^{X}_{\varepsilon_{x}}(x). \end{equation*} The set \begin{equation*} Y := \bigcup_{x \in B}U^{X}_{\varepsilon_{x}}(x) \end{equation*} is open in $X$ because it's the union of open sets in $X$. Furthermore we have $B = A\cap Y$.
But I am having trouble visualizing this: I can't think of a way to sketch the scenario so that $U^{A}_{\varepsilon_{x}}(x) \neq U^{X}_{\varepsilon_{x}}(x)$, since when I draw $X$ to be a box and $A$ a circle inside it and $B$ a smaller circle in $B$, we alway have $U^{A}_{\varepsilon_{x}}(x) = U^{X}_{\varepsilon_{x}}(x)$.
Any help is greatly appreciated :)
Draw $X$ to be the plane and $A$ to be the $x$-axis.