I know same question has an answer already. But I'm rather doubtful if the proof given there actually proves the claim for finite cardinalities.
For infinite cardinals $\kappa (=w(X) \;\;\text{in the proof})$ we do have that, say, union of $\kappa$ copies of sets of cardinality $\kappa$ is $\kappa$, but of course this isn't necessarily true if $\kappa$ is finite.
So I'm wondering how should one account for cases when $\kappa$ is finite?
Any helps are appreciated
Suppose $\mathcal{B}$ is a finite basis for a space $X$. Then I claim that $X$ actually has a smallest basis, i.e. a basis contained in every other basis.
First, note that for any $x\in X$, there is a smallest open set $U_x$ containing $x$: just take the intersection of all the elements of $\mathcal{B}$ that contain $x$ (here is where we use finiteness of $\mathcal{B}$). This $U_x$ is contained in every open set containing $x$, so if $\mathcal{C}$ is a basis for $X$, $U_x$ must be an element of $\mathcal{C}$ (since $\mathcal{C}$ must contain an open subset of $U_x$ containing $x$).
Now define $\mathcal{B}_0=\{U_x\}_{x\in X}$. First, $\mathcal{B}_0$ is a basis: if $U$ is open and $x\in U$, then $U_x\subseteq U$, and $U_x\in\mathcal{C}$. Moreover, we have shown that any other basis $\mathcal{C}$ contains $\mathcal{B}_0$.
So $\mathcal{B}_0$ is the smallest basis of $X$. In particular, all bases contain a basis of minimal possible cardinality (namely, $\mathcal{B}_0$).
Finite spaces are a special case of spaces having a subset-minimal base: if a space has a subset-minimal base $\mathscr{B}$, and $\mathscr{B}'$ is any base for the space, then $\mathscr{B}\subseteq\mathscr{B}'$.
Suppose that a space $\langle X,\tau\rangle$ has a subset-minimal base $\mathscr{B}$. Let $B\in\mathscr{B}$ be arbitrary; $\mathscr{B}\setminus\{B\}$ is not a base for $X$. If $B$ were a union of members of $\mathscr{B}\setminus B$, $\mathscr{B}\setminus\{B\}$ would be a base for $X$, so there is an $x_B\in B$ such that the only member of $\mathscr{B}$ containing $x_B$ and and contained in $B$ is $B$ itself. It follows that $B=\bigcap\{U\in\tau:x_B\in U\}$. Thus, for each $B\in\mathscr{B}$ there is an $x_B\in B$ such that $B$ is the smallest open set containing $x_B$. Let $\mathscr{B}'$ be any base for $\tau$, and let $B\in\mathscr{B}$; there must be some $B'\in\mathscr{B}'$ such that $x_B\in B'\subseteq B$, but then $B'=B$. Thus, $\mathscr{B}\subseteq\mathscr{B}'$.
Let $D=\{x_B:B\in\mathscr{B}\}$; clearly $D$ is dense in $X$. It’s easy to see that every finite space has a subset-minimal base, and in that case $D=X$. It is not true in general, however, that $D=X$. For a simple counterexample let $X=\{0\}\cup\{2^{-n}:n\in\omega\}$, for $n\in\omega$ let $B_n=\{0\}\cup\{2^{-k}:k\ge n\}$, and let $\tau=\{\varnothing\}\cup\{B_n:n\in\omega\}$. $\{B_n:n\in\omega\}$ is a subset-minimal base, and for each $n\in\omega$ the smallest open set containing $2^{-n}$ is $B_n$, but $0\notin D$, and $0$ has no smallest open nbhd. This shows that a topology with a subset-minimal base need not be an Alexandrov topology, though the space must have a dense subset that is Alexandrov-discrete.
I should note that a space with an Alexandrov-discrete dense subset need not have a subset-minimal base: the space $X$ of the previous paragraph with the topology that it inherits from $\Bbb R$ is a counterexample.
