Here are the definitions:
Fréchet-Urysohn space: A topological space $ X $ where for every $ A \subseteq X $ and every $ x \in \text{cl}(A) $, there exists a sequence $ (x_{n})_{n \in \mathbb{N}} $ in $ A $ converging to $ x $.
How to show that first countable spaces are Fréchet-Urysohn Space?
Let $X$ be a first countable topological space. Let $ x \in \text{cl}(A) $ that means $x$ is a limit point of $A$. By first countability there exists a countable neighborhood base for the point $x$ say $\{N_i\}_{i=1}^{\infty}$ . Now for $x\in N_1$ there exists $x_1\in A$ such that $x_1\in N_1$ , similarly for $x\in N_k$ there exists $x_k\in A$ such that $x_k\in N_1$ .
Claim:$\{x_k\}_{k\in \mathbb{N}}$ converges to $x$. Let $U$ be an arbitrary open neighborhood of $x$ then there exists an $N_i$ such that $N_i\subset U$. Then how to proceed? Maybe this procedure will not work.
Best Answer
This is T000183 in the pi-Base; note that it does not reverse as the Finite complement topology on the real numbers and Fort Space on the Real Numbers are counterexamples.
As noted by Oliver in the comments, to finish your proof, simply assume $N_{n+1}\subseteq N_n$ in your countable base; then for any neighborhood $U$ of $x$, pick $N_n\subseteq U$, and then $\{x_m:m\geq n\}\subseteq U$, showing convergence.