I define topological manifold as in Munkres: Hausdorff, second countable, and every point has a neighborhood homeomorphic to an open subset of Euclidean space. My definition of regular is as in Munkres (i.e. needs to be $T_1$).
A typical proof that a topological manifold being regular, for example as found on another stackexchange page or this external page, goes as follows:
Let $X$ be a topological manifold. (First, we have $T_1$ by Hausdorffness.) Let $x\in X$, and $U$ an open neighborhood of $x$. We need to show that there is a neighbourhood $V$ of $x$ such that $cl_X (V) \subseteq U$, where I explicitly use $cl_X$ to denote that the closure is with respect to $X$ (this will be important later).
If $U$ is not already a neighborhood $x$ that is homeomorphic to an open subset of Euclidean space, intersect it with such a neighborhood of $x$, so WLOG assume $U$ is as such. Let $h:U\to h(U)$ be such a homeomorphism. Now, Euclidean space is metrizable hence regular, so there exists $V\subseteq U$ such that $h(x)\in h(V)$ and $cl_{\mathbb{R}^m}h(V)\subseteq h(U)$. The best I can then conclude is
$$cl_U(V)\subseteq U$$
which is basically saying nothing. The homeomorphism $h$ is only between $U$ and $h(U)$, so I do not see how we know anything about the behavior in the larger space $X$ (in particular, closure with respect to $X$).
In summary: I understand these proofs when they choose $V$, but I do not see why $cl_X (V) \subseteq U$ (the key being the closure is with respect to $X$).
One must make sure that the closure of the neighbourhood of $h(x)$ in $h(U)$ corresponds to a closed set in $X$. The easiest way to achieve that is - I think - compactness. Since $\mathbb{R}^n$ is locally compact, there is a compact $K\subset h(U)$ such that $\overset{\Large\circ}{K}$ is a neighbourhood of $x$. Then $h^{-1}(K)$ is a compact, hence closed subset of $X$, contained in $U$, and thus $V := \overbrace{h^{-1}(K)}^{\Large\circ}$ is a neighbourhood of $x$ with $$\operatorname{cl}_X(V) \subset h^{-1}(K) \subset U.$$