TITLE: Proof for "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."
QUESTION [8 upvotes]: The titular statement is used in the explanation of this answer from several years back. I ran across it while puzzling my way through this text, which I bought while I was still in high school and had no idea that I would probably wish I had spent the money on a better text.
I have found no proof for this statement in my text, nor through Googling (I don't know what to call it other than "minimal basis theorem"), and though I've attempted to prove it myself, I'd like to find a more officially-vetted proof, since I'm self-teaching. 
My (probably terrible) proof: 
Impose a partial ordering on bases of a topological space X, such that $\mathcal{B}_1 \leq \mathcal{B}_2$ if $\mathcal{B}_1 \subseteq \mathcal{B}_2$. For any two which are related by $\mathcal{B}_1 \leq \mathcal{B}_2$ , $|\mathcal{B}_1| \leq |\mathcal{B}_2|$. Now the cardinality for all bases is bounded below since only the trivial space has 1 basis element. 
I get stuck about here, and this is my attempt at resolving:
Using the axiom of choice (?) there must be some set S with infimal cardinality |S| such that $|S| \leq |\mathcal{B}_k|$ for all bases that are subsets of $\mathcal{B}$ but this is not the case for any greater cardinality than |S|. 
Are cardinalities well-ordered in a way that would allow me to conclude that the minimal basis must therefore exist??
Seeing as I have no professor to go and bother about this, and I've only had training in naive set theory when I was in university, I'd love some feedback.

REPLY [3 votes]: The argument for Case 2 in @vow lacks forte's answer seems just a bit too complicated. Here's how I would present it.
Assume that $\mathcal D$ is a base of minimum cardinality, $|\mathcal D|=w(X)\ge\aleph_0.$ Let $\mathcal B$ be any base.
Let $\mathcal P=\{(U,V)\in\mathcal D\times\mathcal D:\text{ there exists } B\in\mathcal B\text{ such that } U\subseteq B\subseteq V\}.$
For each pair $(U,V)\in\mathcal P$ choose $B_{U,V}\in\mathcal B$ with $U\subseteq\mathcal B_{U,V}\subseteq V.$
Let $\mathcal B^*=\{B_{U,V}:(U,V)\in\mathcal P\}\subseteq\mathcal B.$
Then $|\mathcal B^*|\le|\mathcal P|\le|\mathcal D\times\mathcal D|=w(X)^2=w(X).$
To see that $\mathcal B^*$ is a base, consider any point $x\in X$ and any neighborhood $W$ of $x.$ Choose $V\in\mathcal D$ with $x\in V\subseteq W;$ then choose $B\in\mathcal B$ with $x\in B\subseteq V\subseteq W;$ then choose $U\in\mathcal D$ with $$x\in U\subseteq B\subseteq V\subseteq W.$$ Then we have $(U,V)\in\mathcal P,\ B_{U,V}\in\mathcal B^*,$ and $$x\in U\subseteq B_{U,V}\subseteq V\subseteq W.$$