Suppose $X$ is a topological space that is locally euclidean of dimension some $n \in \Bbb{N}$. Show that $X$ is first countable.
My attempt: Let $p\in X$ and $U$ a neighborhood of $p$. By assumption, there exists a neighborhood $U'$ of $p$ such that $U'$ is homeomorphic to a first countable space. Hence $U'$ is first countable. Since $U\cap U' \subseteq U'$. It follows that $U \cap U'$ is first countable. Hence each $p \in U\cap U'$ has a local basis, $\mathbb{B}_p$. Since $U\cap U'$ is open in $U'$ and $U'$ is open in $X$, each term in $\mathbb{B}_p$ is open in $X$. Hence $\mathbb{B}_p$ the required local basis.
Is my attempt correct? What would be a better proof?
Almost. You forgot to mention countability at a later point in the proof ("has a local basis" instead of "has a countable local basis"). You don't need to start with $U$, that only distracts.
Could again be more direct: let $p \in X$ and $U$ its locally Euclidean open neighbourhood, so there is a homeomorphism $h: U \to \Bbb R^n$. $\Bbb R^n$ is first countable and so $U$ is too. Hence $p$ has a countable local base $\Bbb B_p$ in $U$, but as $U$ is open, all members of this local base are open in $X$ too, and moreover form a local base at $p$ for $X$: if $O$ (open in $X$) contains $p$, there is a member $B \in \Bbb B_p$ with $B \subseteq O \cap U \subseteq O$ ($O \cap U$ is open in $U$ and we apply the fact that $\Bbb B_p$ is a local base for $U$). As $p$ is arbitrary, $X$ is first countable.