Why $X$ contains a countable $\pi$-basis? I don't understand the following statement.
First, I write what a $\pi$-bases means. Let $X$ be a topological space and $\mathcal{B}$ a family of non-empty open sets. We call $\mathcal{B}$ a $\pi$-basis if for every non-empty open $U\subseteq X$, there exists $B\in\mathcal{B}$ such that $B\subseteq U$.
Now, we have a Tychonoff countable space $X$, such that it contains a discrete countable subspace. Then we would have that $X$ contains a countable $\pi$-basis.
Does anyone see why this must be true?
Thanks.
 A: It’s not true. Let $D=\{0,1\}$ have the discrete topology, and let $Y={^{\Bbb R}D}$, the set of functions from $\Bbb R$ to $D$, with the product topology. If $F=\{x_1,\ldots,x_n\}$ is a finite subset of $\Bbb Q$, $x_1<\ldots<x_n$, and $\varphi$ is a function from $\{0,\ldots,n\}$ to $D$, define $f_{F,\varphi}\in X$ as follows:
$$f_{F,\varphi}(x)=\begin{cases}
\varphi(0),&\text{if }x<x_1\\
\varphi(k),&\text{if }x_k\le x<x_{k+1}\text{ for some }k<n\\
\varphi(n),&\text{if }x\ge x_n\;.
\end{cases}$$
Let $X=\big\{f_{F,\varphi}\in Y:F\subseteq\Bbb Q\text{ is finite and }\varphi\text{ is a function from }\{0,\ldots,|F|\}\text{ to }D\big\}$; $X$ is a countable, dense subspace of $Y$. $Y$ is Tikhonov, so $X$ is Tikhonov as well.
For $n\in\Bbb Z$ let $g_n:\Bbb R\to D$ be the indicator function of $[n,n+1)$; clearly $g_n\in X$, and it’s not hard to check that $\{g_n:n\in\Bbb Z\}$ is a closed, discrete subset of $X$.
For each finite $F\subseteq\Bbb R$ and $\varphi:F\to D$ let $B(F,\varphi)=\{f\in X:f\upharpoonright F=\varphi\}$, and let 
$$\mathscr{B}=\left\{B(F,\varphi):F\subseteq\Bbb R\text{ is finite and }\varphi\in{^FD}\right\}\;;$$
$\mathscr{B}$ is a base for $X$.
Now suppose that $\mathscr{U}=\{U_n:n\in\omega\}$ is a countable family of non-empty open sets in $X$. For each $n\in\omega$ choose a $B(F_n,\varphi_n)\in\mathscr{B}$ such that $B(F_n,\varphi_n)\subseteq U_n$, and let $\mathscr{B}_0=\{B(F_n,\varphi_n):n\in\omega\}$. Clearly $\bigcup_{n\in\omega}F_n$ is countable, so we can choose $x\in\Bbb R\setminus\bigcup_{n\in\omega}F_n$. Let $\varphi:\{x\}\to D:x\mapsto 0$; then for each $n\in\omega$ we have
$$B(F_n,\varphi_n)\nsubseteq B\big(\{x\},\varphi\big)\;,$$
since each $B(F_n,\varphi_n)$ contains points $f\in X$ such that $f(x)=1$. Thus, $\mathscr{U}$ is not a $\pi$-base for $X$, and $X$ has no countable $\pi$-base.
