Direct Sum of Topological $R$-modules

Question

Please help me. I am proving the following property: Let $R$ be a topological ring and $\{A_{\lambda}\}_{\lambda\in\Lambda}$ be a family of topological $R$-modules. Let $B_{\lambda}$ be an untopologized $R$-module and $A_{\lambda}$ be a submodule of $B_{\lambda}$ for every $\lambda\in\Lambda$. For every $\lambda\in\Lambda$, we extend the topology of $A_{\lambda}$ to $B_{\lambda}$ by taking as a basis for the neighborhoods of $0$ in $B_{\lambda}$ the neighborhoods of $0$ in $A_{\lambda}$. The direct sums of $\{A_{\lambda}\}_{\lambda\in\Lambda}$ and $\{B_{\lambda}\}_{\lambda\in\Lambda}$ are $\underline{A}=\bigoplus_{\lambda\in\Lambda}A_{\lambda}$ and $\underline{B}=\bigoplus_{\lambda\in\Lambda}B_{\lambda}$ with topology on $\underline{A}$ is $\tau_{coprod}$ and topology on $\underline{B}$ is $\tau'_{coprod}$. Topology $\tau_{coprod}$ is the supremum of all the $R$-module topologies on $\underline{A}$ such that for all $\lambda\in\Lambda$, $\varepsilon_{\lambda}\colon A_{\lambda}\to \bigoplus_{\lambda\in\Lambda}A_{\lambda}$ is continuous and the same way applies to $\tau'_{coprod}$. Furthermore, $\tau'_{coprod}\mid_{\bigoplus_{\lambda\in\Lambda}A_{\lambda}}=\tau_{coprod}$. We construct a commutative diagram: $$\require{AMScd} \begin{CD} A_{\lambda} @>{\varepsilon_{\lambda}}>> \bigoplus_{\lambda\in\Lambda}A_{\lambda}\\ @V{i}VV @V{i'}VV \\ B_{\lambda} @>{\varepsilon'_{\lambda}}>> \bigoplus_{\lambda\in\Lambda}B_{\lambda} \end{CD}$$ Since the topology on $B_{\lambda}$ is an extension of topology on $A_{\lambda}$ for every $\lambda\in\Lambda$, then $A_{\lambda}$ is an open set of $B_{\lambda}$. This implies that $i$ is an embedding. I am trying to prove that $i'$ is an embedding by proving $\bigoplus_{\lambda\in\Lambda}A_{\lambda}$ is an open set of $\bigoplus_{\lambda\in\Lambda}B_{\lambda}$. Let $X$ be an open set in $\bigoplus_{\lambda\in\Lambda}A_{\lambda}$. Since $\varepsilon_{\lambda}$ is continuous, then $\varepsilon^{-1}_{\lambda}(X)$ is an open set in $A_{\lambda}$. On the other hand, $i$ is an embedding, then $i(\varepsilon^{-1}_{\lambda}(X))$ is an open set in $B_{\lambda}$. If I take $i(\varepsilon^{-1}_{\lambda}(X))$ to $\varepsilon'_{\lambda}$, then $\varepsilon'_{\lambda}(i(\varepsilon^{-1}_{\lambda}(X)))=(\varepsilon'_{\lambda}i)(\varepsilon^{-1}_{\lambda}(X))=(i'\varepsilon_{\lambda})(\varepsilon^{-1}_{\lambda}(X))=i'(X)$, but $\varepsilon'_{\lambda}$ is not an open map. Is my proof correct? What conditions should I add to make $\bigoplus_{\lambda\in\Lambda}A_{\lambda}$ an open set $\bigoplus_{\lambda\in\Lambda}B_{\lambda}$. Thank you for any help.