Suppose $\{\mathcal{T}_{\alpha}\mid\alpha\in J\}$ is a family of topologies on $Y$ such that each inclusion map $j_{\alpha}\colon(Y,\mathcal{T}_{\alpha})\hookrightarrow X$ is continuous. Let $\mathcal{T}$ be the intersection $\bigcap_{\alpha\in J}\mathcal{T}_{\alpha}$ Observe that $\mathcal{T}$ is also a topology on $Y$ Let $U$ be open in $X$ By continuity of $j_{\alpha}$ the set $j_{\alpha}^{-1}(U)=j^{-1}(U)$ is open in each $\mathcal{T}_{\alpha}$ consequently, $j^{-1}(U)$ is also in $\mathcal{T}$ This shows that there is a weakest topology on $Y$ making inclusion continuous.

We claim that any topology strictly weaker than $\mathcal{S}$ fails to make the inclusion map continuous. To see this, suppose $\mathcal{S}_0\subsetneq\mathcal{S}$ is a topology on $Y$ Let $V$ be a set open in $\mathcal{S}$ but not in $\mathcal{S}_0$ By the definition of subspace topology, $V=U\cap Y$ for some open set $U$ in $X$ But $j^{-1}(U)=V$ which was specifically chosen not to be in $\mathcal{S_0}$ Hence $\mathcal{S_0}$ does not make the inclusion map continuous. This completes the proof.