Existence of subbasis with cardinality of $X$ Let $S$ be a subbasis for a topology on infinite set $X$.Does there  exist any subset $H$ of $S$ with $Card(H)\leq Card(X)$ such that $H$ be a subbasis of that topology on $X$ $?$proof it or give me a counterexample please.
 A: Let $Y=\Bbb N\times\Bbb N$, let $p$ be a point not in $Y$, and let $X=\{p\}\cup Y$. For each function $f:\Bbb N\to\Bbb N$ let
$$B_f=\{p\}\cup\{\langle k,\ell\rangle\in Y:\ell\ge f(k)\}\;.$$
Topologize $X$ by making each point of $Y$ isolated and taking $\left\{B_f:f\in{^{\Bbb N}\Bbb N}\right\}$ as a local base at $p$.
To show that this is a counterexample it suffices to show that $X$ has no countable subbase.
Suppose that $\mathscr{S}$ is a subbase for $X$. Let 
$$\mathscr{B}=\left\{\bigcap\mathscr{F}:\mathscr{F}\subseteq\mathscr{B}\text{ and }\mathscr{F}\text{ is finite}\right\}\;,$$
the base generated by $\mathscr{S}$. $\mathscr{S}$ is infinite, so $\mathscr{S}$ has $|\mathscr{S}|$ finite subsets, and therefore $|\mathscr{B}|=|\mathscr{S}|$. If $\mathscr{S}$ were countable, $\mathscr{B}$ would also be countable, and $X$ would be second countable and hence first countable. But it’s easy to see that it has no countable local base at $p$.
Suppose that $\mathscr{U}=\{U_n:n\in\Bbb N\}$ is a countable family of open nbhds of $p$. For each $n\in\Bbb N$ there is an $f_n:\Bbb N\to\Bbb N$ such that $B_{f_n}\subseteq U_n$. Define 
$$g:\Bbb N\to\Bbb N:k\mapsto 1+\max\{f(\ell):\ell\le k\}\;;$$
then $B_g\nsupseteq B_{f_n}$ for each $n\in\Bbb N$, so $\mathscr{U}$ is not a local base at $p$.
