This is part of Lee's Introduction to Topological Manifolds exercise 3.12. While the question contains multiple pieces, I am primarily interested in solving the following:
Suppose S is a subspace of the topological space X. Show that every subspace of a first countable space is first countable
I attempted to do the following:
Consider $s \in S$, and let $U \subseteq S$ be an open set that contains $s$. Since $S$ is a subspace topology, we can write $U$ as $U = S \cap V$, where $V\subset X$ is open in $X$. Clearly $s \in V$. Since $X$ is first countable, we know that there exists $B \in \mathcal{B}_s^X$ such that $B \subseteq V$ and $s \in B$.
But I do not see how I can continue my proof. Ideally, I would like to demonstrate that this somehow leads to the element $s$ lying in a collection of open sets such that $U$ contains one of those sets.
Any recommendations would be appreciated
Best Answer
Let $p\in S$ and let $(V_n)_{n\in\Bbb N}$ be a fundamental system of neighboroods of $p$ in $X$. Then $(V_n\cap S)_{n\in\Bbb N}$ is a fundamental system of neighborhoods of $p$ in $S$.