I wrote up this proof for practice, if someone could critic it, thank you!
Let $T$ be a topology on a set $X$. Let $T_{A}$ denote a topology on the subspace $A$.
Claim. If $A \subset X$ is $T$-open, any $T_{A}$-open set is also $T$-open.
Proof. Assume $A \subset X$ is $T$-open. Then $A \in T$. Let's denote the subspace topology as the following:
$T_{A} = \{B \cap A \vert B \in T \text{ and } A \in T\}$.
Let $C_{n}$ be any $T_{A}$-open set. Then $C_{n} = \{C_m \cap A \text{ where } C_m \in T \text{ and } A \in T\}$ For any $C_n \in T_A$ where $n \in \mathbb{N}$, there exists a $C_m \in T$ where $m \in \mathbb{N}$ such that $C_n = C_m \cap A$ for $A \in T$. Since $A \in T$ and $C_m \cap \in T$, the intersection of two $T$-open sets is $T$-open; hence $C_n \in T$. Hence, any $T_A$ open set is also $T$-open.
I wasn't sure if I could prove this by induction. Thank you for reading!
This has nothing to do with induction or countable index sets:
Suppose $O$ is $\mathcal{T}_A$-open, which means that $O = O' \cap A$ where $O' \in \mathcal{T}$.
But then $O$ is the intersection of two sets in $\mathcal{T}$ (namely $O'$ and $A$ (by asumption)), and hence in $\mathcal{T}$ itself, as topologies are closed under finite intersections.
Hence $\mathcal{T}_A \subseteq \mathcal{T}$.