If $(X, \tau)$ is $T_1$ and there is an open-closed set basis for the $X$ topology, show that $X$ is completely regular.
we will first remember some definitions completely regular: A topological space $X$ is called completely regular, if given a closed $A$ in $X$ and a point $b \in X \setminus A$, there exists a continuous function $f: X → [0, 1]$ such that $f (A) \subseteq \{0\}$ and $f (b) = 1$. All normal space is completely regular.
Let $X$ be a topological space. $X$ is said to be a $T_1$ space if Given $x$ and $y$ in $X$ with $x \neq y$, there exist $U \in U_x$ and $V \in U_y$ such that $y \notin U$ and $x \notin V$.
I think it would be easier to test if it is possible that it is normal, but I don't know how to use the hypothesis of the closed and open base, any help, please?
Let $x\in X$ be any point and $F\not\ni x$ be any closed subset of $X$. Since $X$ has a basis consisting of clospen (tht is, closed and open) sets, there exists a clopen set $U$ such that $x\in U\subset X\setminus F$. Then a function $f:X\to [0,1]$ such that $f(U)\subset \{0\}$ and $f(X\setminus U)\subset \{1\}$ is continuous and separates $x$ and $F$.