Given a locally Euclidean space X of dimension n and a point $p \in X$, by definition there exists a neighborhood $U \subset X$ and a homeomorphism $\phi$ such that $\phi (U)$ is an open subset in $\mathbb{R}^n$
I wanted to show that every locally Euclidean space is first countable. So for $p \in X$ arbitary, take a chart $(U, \phi)$ and use that $\mathbb{R}^n$ is first countable to get a countable neighborhood basis $\{U_i\}$ of $\phi(p)$. Take an arbitary neighborhood N of p. I would now like to use that $\phi(N)$ is also a neighborhood of $\phi(p)$. And then use first countability of $\mathbb{R}^n$ to show that there exists a $U_j \subset \{U_i\}$ so that $\phi^{-1}(U_j) \subset N$
If we now look at an arbitary neighborhood of p, does it follow that $\phi$ is also a homeomorphism on N, so that $\phi(N)$ is also a neighborhood of $\phi(p)$ or is $\phi$ only defined on U?
The space $X$ is locally Euclidean of dimension $n$ i.e. each $x\in X$ has a open nbd $U_x$ and a homeomorphism $\phi_x:U_x\rightarrow \phi_x(U_x)$ such that $\phi_x(U_x)$ is open in $\Bbb R^n$.
Choose a countable local base $\{V_n\}_n$ at $\phi_x(x)$, this is possible as $\Bbb R^n$ is first countable , then $\{V_n\cap \phi_x(U_x)\}_n$ is also countable local base at $\phi_x(x)$. We claim that $\{\phi_x^{-1}(V_n\cap \phi_x(U_x))\}_n$ is a countable local base at $x$. Certainly each $ \phi_x^{-1}(V_n\cap \phi_x(U_x))$ is open in $X$ containing $x$ since $\phi_x^{-1}$ is also continuous.
Now choose a open nbd $N_x$ of $x$ then $U_x\cap N_x$ is a open nbd of $x$. So that $\phi_x(U_x\cap N_x)(\subseteq \phi_x(U_x))$ is open nbd of $\phi_x(x)$ as $\phi_x$ is open map. Therefore $\phi_x(U_x\cap N_x)$ contains some $V_m \cap \phi_x (U_x)$ i.e. $N_x\supseteq U_x\cap N_x=\phi_x^{-1}(\phi_x(U_x\cap N_x))\supseteq \phi_x^{-1}(V_m \cap \phi_x (U_x))$. Hence we are done.