Let $X$ be a topological vector space. Let $\mathcal B$ be a local base at $0.$ Then the following statements hold $:$
$(1)$ $\mathcal U, \mathcal V \in \mathcal B \implies \exists\ \mathcal W \in \mathcal B$ such that $\mathcal W \subset \mathcal U \cap \mathcal V.$
$(2)$ $x \in \mathcal U \in \mathcal B \implies \exists\ V \in \mathcal B$ such that $x + \mathcal V \in \mathcal U.$
$(3)$ $\mathcal V \in \mathcal B, x \in X \implies \exists\ 0 \neq \lambda \in \Bbb C\ (\text {or}\ \Bbb R)$ 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 $|\lambda| \lt 1.$
$(1)$ $:$ It follows from the fact that $\mathcal U \cap \mathcal V$ is an open neighbourhood of $0.$ Now since $\mathcal B$ is a local base at $0$ it follows that there exists $\mathcal W \in \mathcal B$ such that $\mathcal W \subseteq \mathcal U \cap \mathcal V.$
$(2)$ $:$ Since $X$ is a topological vector space the map $f_x : X \longrightarrow X$ defined by $y \mapsto (y - x)$ is a homeomorphism and hence open. Thus $\mathcal U - x$ is an open neighbourhood of $0.$ So there exists $\mathcal V \in \mathcal B$ such that $\mathcal V \subseteq \mathcal U - x \iff x + \mathcal V \subseteq \mathcal U.$
$(3)$ $:$ Since $X$ is a topological vector space the map $g_x : \Bbb C \longrightarrow X$ defined by $\lambda \mapsto \lambda x$ is continuous. In particular $g_x$ is continuous at $0.$ Hence there exists $\varepsilon \gt 0$ such that for all $\lambda \in \Bbb C$ with $|\lambda| \lt \varepsilon$ we have $\lambda x \in \mathcal V.$ Now take any $\lambda \in \Bbb C$ with $0 \lt |\lambda| \lt \varepsilon$ and then we are through.
$(4)$ $:$ Since $X$ is a topological vector space the binary operation $(x,y) \mapsto (x+y)$ is continuous. Using the continuity of the binary operation at $(0,0)$ we get open neighbourhoods $V_1$ and $V_2$ of $0$ such that $V_1 + V_2 \subseteq \mathcal U.$ Find $\mathcal W_1, \mathcal W_2 \in \mathcal B$ such that $\mathcal W_1 \subseteq V_1$ and $\mathcal W_2 \subseteq V_2.$ Then $\mathcal W_1 \cap \mathcal W_2$ is an open neighbourhood of $0.$ So there exists $\mathcal V \in \mathcal B$ such that $\mathcal V \subseteq \mathcal W_1 \cap \mathcal W_2.$ Then by the construction $\mathcal V \subseteq V_1 \cap V_2$ and hence $\mathcal V + \mathcal V \subseteq V_1 + V_2 \subseteq \mathcal U.$
Can anybody please help me proving the last statement i.e. $(5)\ $? I am having hard time in proving this. Also please check my arguments from $(1)$-$(4).$
Thanks in advance.
It is not true under your assumption "Let $\mathcal B$ be a local base at $0$." As Sumanta Das comments, take $X = \mathbb R$ and $\mathcal B = \{(−1/n,2/n)\mid n \in \mathbb N\}$. Then no $U \in \mathcal B$ has the property $(−3/4)U \subset U$ although $|−3/4|<1$.
You can show that there exists a local base $\mathcal B'$ at $0$ with properties $(1)−(5)$. Given any $\mathcal B$, for each $U \in \mathcal B$ define $U' = \bigcup_{|λ| \le 1} λU$ and let $\mathcal B'$ be the collection of these $U'$.