First countable space Theorem: A subspace of a first countable space is first countable.
My proof:
Let $X$ be a first countable space. So for each p $\in X$, there exists a countable neighborhood basis for $X$ at $p$. Let $A$ $\subseteq X$ be a subspace and $p\in A$. Since $p$ is a member of the space $X$, let $\mathbb{B}_p$ be the countable neighborhood basis $\mathbb{B}_p$ for $X$ at $p$. Consider $\mathbb{B}_p'$ $=$ $\{$ $A\cap B$ $:$ $B\in \mathbb{B}_p$ $\}$. If $\mathbb{B}_p$ is finite, then so is $\mathbb{B}_p'$ , hence it would be countable. If $\mathbb{B}_p'$ was countably infinite, then define a function $f:$ $\mathbb{B}_p$ $\rightarrow \mathbb{B}_p'$ to be $f(B_i)=A\cap B_i$. Clearly the function is well defined and a surjection; hence, since $\mathbb{B}_p$ is countably infinite, $\mathbb{B}_p'$ is countable. Let $U$ be a neighborhood of $p$ in the subspace topology, so $U$= $A\cap U'$, where $U'$ is a neighborhood of $p$ at $X$ containing some $B\in \mathbb{B}_p$. Hence, $A\cap B \subseteq A\cap U' = U$. Since $A$ is an arbitrary subspace of $X$, every subspace of a first countable space is first countable.
Is it correct?

*

*You don't have to distinguish countable from finite, it's superfluous.

*Of course $\Bbb B'_p$ is by definition the image of $\Bbb B_p$ under $U \to U \cap A$ so is also countable (an image of a countable set is countable).
The proof itself is correct. Don't try to be "hypercorrect", though.
