A misinterpretation
I think you have misunderstood the stated fact in the linked answer. When it is stated that
given any basis of a topological space, you can always find a subset of that basis which itself is a basis, and of minimum possible size.
what is meant is the following
given any basis of a topological space, you can always find a subset of that basis which itself is a basis, and of minimum possible size among all bases.
Note the difference between this interpretation and
given any basis of a topological space, you can always find a subset of that basis which itself is a basis, and of minimum possible size among all bases which are subsets of the given one.
Silly example to illustrate the difference
For this (I am certain correct) interpretation of the statement your given proof will not work. Note, in particular, that nowhere do you essentially use the fact that the collections you have chosen are bases for a topological space. And, if true, it should carry on to any other collection. Note, too, that in that answer it is claimed that this fact together with the second-countability of the space would allow you to get a countable base which is a subset of $\mathcal B$. A priori it would not be known that the minimum cardinality among all bases which are subsets of $\mathcal B$ is $\leq \aleph_0$.
Example. Call an infinite subset $A \subseteq \mathbb R$ halatinous if it either contains a rational number, or is uncountable.
Note that the minimum size of a halatinous subset of $\mathbb R$ is $\aleph_0$. Clearly every halatinous subset of $\mathbb R$ has cardinality at least $\aleph_0$, and $\mathbb N$ is a halatinous subset of $\mathbb R$ of that cardinality.
Note, too, that $A = \mathbb R \setminus \mathbb Q$ is halatinous, however it has no countably infinite halatinous subsets. (The smallest cardinality of a halatinous subset of $A$ is $\aleph_1 > \aleph_0$.)
So while there are countable halatinous subsets of $\mathbb R$, there are also halatinous subsets of $\mathbb R$ with no countable halatinous subsets.
Your proof, had it worked for bases would have equally worked for halatinous subsets of $\mathbb R$, which is impossible. So in order to prove the intended statement you are going to have to make use of the fact that you are starting with a base for a topological space, and are looking for another base.
Sketch of the proof
So let's make a very precise statement. Given a topological space $X$, the weight of $X$, denoted $w(X)$, is defined to be the least cardinality of a base for $X$. (Often $w(X)$ is restricted to take only infinite values, but for the purposes of what follows I'll allow finite values, too.)
Then a more precise statement of the given fact is
Given any basis $\mathcal B$ of a topological space $X$ you can always find a subset $\mathcal B^* \subseteq \mathcal{B}$ which is also a base for $X$, and which satisfies $|\mathcal B^* | = w(X)$.
So let $X$ be a topological space, and let $\mathcal{B}$ be any base. We show that some subset $\mathcal B^* \subseteq \mathcal B$ of cardinality $w(X)$ is also a base for $X$. To start, fix a base $\mathcal D$ for $X$ of cardinality $w(X)$. There are essentially two cases.
Case 1. Assume that $w(X) < \aleph_0$, so in particular $\mathcal D$ is finite.
- Given $x \in X$, $U_x = \bigcap \{ V \in \mathcal D : x \in V \}$ is the smallest open neighborhood of $x$.
- Each $U_x$ is contained in every base for $X$.
- The only sets in $\mathcal D$ are the $U_x$ for $x \in X$. (This uses the fact that $\mathcal D$ is a base of minimum finite size.)
- Therefore $\mathcal D \subseteq \mathcal B$.
Case 2. Assume that $w(X) \geq \aleph_0$. (So that every base for $X$ is infinite.) Then we can proceed as follows.
- Show that given any open $W \subseteq X$ there is a $\mathcal B_W \subseteq \mathcal B$ such that $| \mathcal B_W | \leq w(X)$ and $\bigcup \mathcal B_W = W$. (There is a $\mathcal B^\prime _W \subseteq \mathcal B$ such that $W = \bigcup \mathcal B^\prime_W$. Take $\mathcal D_W = \{ V \in \mathcal D : (\exists U \in \mathcal B^\prime_W )( V \subseteq U ) \}$. Note that $| \mathcal D_W | \leq w(X)$ and $\bigcup \mathcal D_W = W$. For each $V \in \mathcal D_W$ fix $U_V \in \mathcal B^\prime_W$ such that $V \subseteq U_V$. Take $\mathcal B_W = \{ U_V : V \in \mathcal D_W \}$.)
- Take $\mathcal B ^* = \bigcup \{ \mathcal B _V : V \in \mathcal D \}$.
- Since each set in $\mathcal B^*$ is open, and each set in the base $\mathcal D$ is a union of a subfamily of $\mathcal B^*$, it follows that $\mathcal B^*$ is a base for $X$. (Therefore $w(X) \leq | \mathcal B^* |$.)
- Making use of the fact that $w(X)$ is infinite, $$| \mathcal B ^* | = | \bigcup \{ \mathcal B_V : V \in \mathcal D \} | \leq \sum_{V \in \mathcal D} | \mathcal B_V | \leq \sum_{V \in \mathcal D} w(X) = w(X) \cdot w(X) = w(X).$$
