Topology "generated by" normal subgroups (Topological groups). 
Let $G$ be a group and $L$ be a non-empty family of normal subgroups such that if $K_{1},K_{2} \in L$ and $K_{3}$ is a normal subgroup containing $K_{1} \cap K_{2}$ then $K_{3} \in L$. Let $T$ be a family of unions of sets of cosets $Kg$ with $K \in L,g \in G$. Show that $T$ is a topology in $G$ and that $G$ is a topological group with respect to this topology. Show that $L$ is the set of open normal subgroups of $G$ with respect to this topology.

This is the first problem of Wilson's book "Profinite Groups". I'm having trouble with this question.

Using YCor's comment, I was able to prove that $T$ is a topology with basis:
$$B = \left\{C_{I,J} \mid C_{I,J} = \{K_{i}g_{j}\}_{i \in I, j \in J}\right\}.$$
Now, to prove that $G$ is topological group, I need to show that the maps $\varphi: G \times G \to G$ given by $\varphi(x,y) = xy$ and $\psi: G \to G$ given by $\psi(x) = x^{-1}$ are continuous.
I'm still not familiar with topological groups (this is the first problem of the book), so I'm having a bit of trouble time with this question.
So, I need to show that if $U$ is open in $G$, $\varphi^{-1}(U)$ is open in $G \times G$. But, is enough consider $U \in B$. I worked with a simple subset of $B$: $\{Kg\}$ for some $K \in L$ and $g \in G$. The sets
$$Kg \times \{1\}1$$
$$K1\times \{1\}g$$
$$Kg^{-n}\times \{1\}g^{n+1}$$
$$Kg^{n+1} \times \{1\}g^{-n}$$
and the "symmetrical" products are maps in $\{Kg\}$ and the union of this sets is in $T$. If I'm not completely wrong, I think that the general case is similar, but I dont know how to formalize.
Also, for the last part, Im having troubles to show that the set of open normal sets of $G$ is cointained in $L$. I would like a hint for it.
 A: The assumption on $L$ is equivalent to the following two conditions:
(a) If $K_1, K_2 \in L$, then $K_1 \cap K_2 \in L$. [because $K_1 \cap K_2$ is normal]
(b) If $K \in L$ and $N$ is a normal subgroup such that $K \subset N$, then $N \in L$. [take $K_1 = K_2 = K$]
We shall now prove a few claims.
1) Let $K \in L$ and $g,g' \in G$ such that $g \in Kg'$. Then $Kg'= Kg$.
We have $g = kg'$ for some $k \in K$. Since $Kk' = K$ for any $k' \in K$, we get $Kg' = K k^{-1} g = K g$.
2) Let $K_1, K_2 \in L$ and $g_1,g_2 \in G$. If  $K_1g_1 \cap K_2g_2 \ne \emptyset$, then $K_1g_1 \cap K_2g_2 = (K_1 \cap K_2)g$ for some $g \in G$.
Let $g \in K_1g_1 \cap K_2g_2$. Then $K_1g_1 = K_1g, K_2g_2 = K_2g$. Hence $K_1g_1 \cap K_2g_2 = K_1g \cap K_2g = (K_1 \cap K_2)g$.
3) $T$ is a topology on $G$.
$\emptyset \in T$ (it is the union of the empty family of cosets $Kg$.) 
$X \in T$ ($X = \bigcup_{g \in G} Kg$ for any $K \in L$.)
$T$ contains all unions of members of $T$ (obvious from the definition.)
$T$ contains the intersection of any two members of $T$:
For $k = 1,2$ let $U_k = \bigcup_{i_k \in A_k, j_k \in B_k} K_{k,i_k}g_{k,j_k}$. Then $U_1 \cap U_2 = \bigcup_{i_1 \in A_1, j_1 \in B_1,i_2 \in A_2, j_2 \in B_2} K_{1,i_1}g_{1,j_1} \cap K_{2,i_2}g_{2,j_2}$ which belongs to $T$ by 2).
4) For each $g \in G$, the set $L(g) = \{ Kg \mid \ K \in L \}$ is a base of open neigborhoods of $g$.
Let $U$ be an open neighborhood of $g$. Choose $K \in L$ and $g' \in G$ such that $g \in Kg' \subset U$. By 1) we have $Kg' = Kg$.
5) $\varphi$ is continuous.
Let $g_1,g_2 \in G$ and $U$ be an open neighborhood of $g_1g_2$. Choose $K \in L$ such that $Kg_1g_2 \subset U$. The sets $V_i = Kg_i$ are open neighborhoods of $g_i$. Since $K$ is normal, we have $g_1K = Kg_1$. We get $\varphi(V_1,V_2) = Kg_1Kg_2 = KKg_1g_2 = Kg_1g_2 \subset U$.
6) $\psi$ is continuos.
Let $g \in G$ and $U$ be an open neighborhood of $g^{-1}$. Choose $K \in L$ such that $ Kg^{-1}  \subset U$. The set $V = Kg$ is an open neighborhood of $g$. We get $\psi(V) = (Kg)^{-1} = g^{-1}K^{-1} = g^{-1}K = Kg^{-1} \subset U$. 
7) $L$ is the set of open normal subgroups.
Let $N$ be an open normal subgroup.  It contains the neutral $e$ of $G$. Hence there exists $K \in L$ such $K = Ke \subset N$. By assumption on $L$ we get $N \in L$.
Conversely, if $K \in L$, then $K = Ke$ is open.
