Question: The subspace topology $\tau\_A$ on $A \subset X$ is induced by the inclusion $\iota :A\hookrightarrow X$. Now if the topology is induced from open sets $U \subset X$ by $\iota^{-1}[U] = \{x \in A : \iota(x) \in U\} = \{x \in A : x \in U \} = A \cap U$, then in order for the sets $A \cap U$, where $U$ is any open set in $X$ to be open wouldn’t $A$ must be also?

It’s confusing that in my book they’re just stating that $A \subset X$, but $A$ could be close for all I know and then $A \cap U$ wouldn’t be open?

Answer: 
No, $A$ doesn't have to be an open subset of $X$. We simply say that a subset $B$ of $A$ is open when $B=A\cap U$, for some open subset $U$ of $X$.

For instance $\{x\in\Bbb Q\mid-1<x<1\}$ is an open subset of $\Bbb Q$ with respect to the subspace topology (when we see $\Bbb Q$ as a subset of $\Bbb R$, and we take in $\Bbb R$ its usual topology), since$$\{x\in\Bbb Q\mid-1<x<1\}=(-1,1)\cap\Bbb Q.$$The fact that neither $\Bbb Q$ nor $\{x\in\Bbb Q\mid-1<x<1\}$ are open subsets of $\Bbb R$ is irrelevant here.
This is how you define your open sets in the induced topology. It is perfectly possible that an intersection $A\cap U$ is not open as a subset of $X$. For example, $[0,1)$ is open in the induced topology (from the standard one) on $[0,1]$.