Let $x\in U$ , open subset of the manifold $X$ . There is a neighborhood $V$ of $x$ homeomorphic to an open euclidean space. Then $W=U\cap V$ is open in $V$ and is homeomorphic to an open subset of the euclidean space. Let $h$ be the homeomorphism. There is a neighborhood $h(A)$ of $h(x)$ such that $\overline{h(A)}\subseteq h(W)$ . Then $A$ is a neighborhood of $x$ such that $\overline{A}\subseteq W\subseteq U$ . Now, important! This does not show yet that the space is regular. Consider the following example. Let us take the standard topology on $\mathbb{R}$ and make a copy of the origin, a new point $a$ . The basis for the topology is the collection of open intervals $(a,b)\subseteq\mathbb{R}$ and open intervals containing 0 with $a$ substituted for $0$ : $(c,d)-\{0\}\cup\{a\}$ where $c<0<d$ . Obviously, this space is a manifold (if we do not require the Hausdorff condition in the definition), however, it is not Hausdorff: we can not separate $0$ and $a$ (even though for every point $x$ and its neighborhood $U$ there is a neighborhood $V$ of $x$ such that $\overline{V}\subseteq U$ , i.e. it is "almost regular"). So, we do need require a manifold be Hausdorff to complete our proof that it is regular.