 Solution

Every subspace-open set has the form $A\cap U=i^{-1}(U)$ for some $U\in \mathcal{T}$. Continuity of $i:(A,{\mathcal{T}}^{\prime })\to (X,\mathcal{T})$ makes every such inverse image open in ${\mathcal{T}}^{\prime }$. Hence $\boxed{{\mathcal{T}}_A\subseteq {\mathcal{T}}^{\prime }}$.