Let $X$ be a vector space. Let $\mathcal B$ be a collection of subsets of $X$ containing $0$ satisfying the following properties $:$
$(1)$ $\mathcal U, \mathcal V \in \mathcal B \implies \exists\ \mathcal W \in \mathcal B$ such that $\mathcal W \subseteq \mathcal U \cap \mathcal V.$
$(2)$ $x \in \mathcal U \in \mathcal B \implies \exists\ \mathcal V \in \mathcal B$ such that $x + \mathcal V \subseteq \mathcal U.$
$(3)$ $\mathcal V \in \mathcal B, x \in X \implies \exists\ 0 \neq \lambda \in \Bbb C\ (\text {or}\ \Bbb R)$ such that $\lambda x \in \mathcal V.$
$(4)$ $\mathcal U \in \mathcal B \implies \exists\ \mathcal V \in \mathcal B$ such that $\mathcal V + \mathcal V \subseteq \mathcal U.$
$(5)$ $\mathcal U \in \mathcal B \implies \lambda\ \mathcal U \subseteq \mathcal U,\ \forall\ \lambda$ with $|\lambda| \lt 1.$
Then $\mathcal B_1 = \{x + \mathcal U\ |\ \mathcal U \in \mathcal B \}$ is a base for a topology $\mathscr T$ on $X$ which makes $X$ a topological vector space.
$\textbf {Attempt} :$ First of all it is clear that $$\bigcup\limits_{x \in X,\ \mathcal U \in \mathcal B} (x + \mathcal U) = X.$$ Now let $x + \mathcal U, y + \mathcal V \in \mathcal B_1$ such that $(x + \mathcal U) \cap (y + \mathcal V) \neq \varnothing.$ Let $z \in (x + \mathcal U) \cap (y + \mathcal V).$ Then $(z - x) \in \mathcal U$ and $(z - y) \in \mathcal V.$ So by property $(2)$ it follows that there exist $\mathcal V_1,\mathcal V_2 \in \mathcal B$ such that $(z - x) + \mathcal V_1 \subseteq \mathcal U$ and $(z - y) + \mathcal V_2 \subseteq \mathcal V.$ Let $V' = \mathcal V_1 \cap \mathcal V_2.$ Then by property $(1)$ there exists $\mathcal W \in \mathcal B$ such that $\mathcal W \subseteq V'.$ Then $(z-x) + \mathcal W \subseteq (z - x) + \mathcal V_1 \subseteq U$ and $(z - y) + \mathcal W \subseteq (z - y) + \mathcal V_2 \subseteq \mathcal V.$ In other words $z + \mathcal W \subseteq (x + \mathcal U) \cap (y + \mathcal V).$ This shows that $\mathcal B_1$ is a basis for a topology $\mathscr T$ on $X.$
Now let $x,y \in X.$ Take a basic open set $z + \mathcal U$ containing $x + y,$ for some $z \in X$ and $\mathcal U \in \mathcal B.$ Then $(x + y - z) \in \mathcal U.$ So by property $(2)$ it follows that there exists $\mathcal V \in \mathcal B$ such that $(x + y - z) + \mathcal V \subseteq U$ i.e. $(x + y) + \mathcal V \subseteq z + \mathcal U.$ Now by property $(4)$ there exists $\mathcal V' \in \mathcal B$ such that $\mathcal V' + \mathcal V' \subseteq \mathcal V.$ Now consider the open neighbourhood $(x + \mathcal V') \times (y + \mathcal V') \subseteq X \times X$ of $(x,y).$ Then $(x + \mathcal V') + (y + \mathcal V') \subseteq (x + y) + \mathcal V \subseteq z + \mathcal U.$ This shows that the binary operation $(x,y) \mapsto (x + y)$ is continuous with respect to the topology $\mathscr T$ on $X.$
But I find difficulty to prove the continuity of the scalar multiplication. I tried to proceed as follows $:$
Let $\lambda \in \Bbb C$ and $x \in X.$ Let $z + \mathcal U \in \mathcal B_1$ be a basic open neighbourhood of $\lambda x.$ It implies that $(\lambda x - z) \in \mathcal U.$ Therefore by property $(2)$ there exists $\mathcal V \in \mathcal B$ such that $(\lambda x - z) + \mathcal V \subseteq \mathcal U$ i.e. $\lambda x + \mathcal V \subseteq z + \mathcal U.$ Now I take $0 \lt \varepsilon \leq 1.$ Then take the open neighbourhood $B(\lambda, \varepsilon) \times (x + \mathcal V) \subseteq \Bbb C \times X$ containing $(\lambda,x).$ Can we somehow say that for $\lambda' \in B(\lambda,\varepsilon)$ we have $\lambda' x + \lambda' \mathcal V \subseteq \lambda x + \mathcal V\ $? What I know by property $(5)$ is that $(\lambda' - \lambda) \mathcal V \subseteq \mathcal V.$ Does it help anyway? Also how do I show that this topological space is Hausdorff? Any help in this regard will be highly appreciated.
Thanks in advance.