$A$ is open as a subset of $Y$ $\Leftrightarrow$ it is the intersection with $Y$ of a set which is open in $X$. $A$ is open as a subset of $Y$ $\Leftrightarrow$ it is the intersection with $Y$ of a set which is open in $X$
I was going through this theorem in Baby Rudin(chapter $2$).This didn't seem very intuitive to me so I was trying to understand it's implication in the metric space.

Let $(X,d)$ be a metric space and $A \subset X$.Let $d_A$ be the induced metric space on $A$ . Let $d'$ be the metric on $A$ where the basic open balls are of the form $\{B_d(a;r) \cap A\}$.Then the above theorem claims that the two metrics are equivalent.

Is my interpretation of the theorem correct? How does one see the intuition behind this theorem in any arbitrary topological space?
 A: It's a definition, not a theorem. In all the treatments of topology I've seen, at least.
It's still important to get an intuition for why it's a sensible definition, though! Here's my take on it:
Open sets are a tool to define a sense of nearness. A sequence $a_n\neq a$ gets near to a point $a$ if every open set containing $a$ contains an element of the sequence. A point $a$ is near to the set $S$ (meaning that it's in the closure of $S$) if every open set containing $a$ intersects $S$.
Now the open sets in a subspace should preserve these properties. If $X$ is a subspace of $Y$ and $a\in X, S\subseteq X$ such that $a$ is near $S$ in the topology of $Y$, then $a$ should still be near $S$ in the topology of $X$. If we define open subsets of $X$ as those which are intersections of $X$ with open subsets of $Y$, this need is fulfilled. So it's a reasonable definition.
A: By definition, $B_d(a;r)\cap A=B_{d_A}(a;r)$ for all $a\in A$ and all $r>0$. More generally, if $(X,\tau_X)$ is a topological space, $Y\subset X$, and $\tau_Y=\{U\cap Y:U\in\tau_X\}$, then $(Y,\tau_Y)$ is a topological space, and $\tau_Y$ is called a subspace topology.
