I am working on a proof that given some closed $Y\subseteq X$, where $X$ is a manifold with empty (manifold) boundary and $Y$ is a manifold of the same dimension as $X$ and $X\setminus Y$ is not closed, the topological boundary $\partial_\tau Y$ with respect to the subspace topology endowed by $X$ is equal to the manifold boundary $\partial Y$.
We begin by showing $\partial Y\subseteq \partial_\tau Y$. Take any point in $y\in Y^\circ$. Because this subset of $Y$ is open in $X$, we can construct a locally Euclidean nbhd of $y$ completely contained in $Y^\circ$. Thus, the only points in $Y$ that may not have a locally Euclidean nbhd are those along $\partial_\tau Y$, i.e., $\partial Y\subseteq \partial_\tau Y$.
We now show that $\partial_\tau Y\subseteq \partial Y$. Because we have supposed $Y$ to be closed, $\partial_\tau Y\subseteq Y$. Now, take some $y\in\partial_\tau Y$. We know that there does not exist any open nbhd of $y$ (w.r.t. to the topology of $X$) contained in $Y$ that is homeomorphic to an open subset of $\mathbb{R}^n$ (because $X\setminus Y$ is not closed).
It's at this point that I am struggling with the proof. I need to figure how to construct a homeomorphism for some nbhd of $y$ open w.r.t. the subspace topology endowed by $X$ which maps this nbhd of $y$ to $\mathbb{H}^n$. I imagine this has to do with choosing our nbhd of $y$ s.t. it contains part of a connected component of $\partial_\tau Y$ and then transforming this so that the points have a final component of $0$, thus being a member of $\partial\mathbb{H}^n$. However, I am unsure how I would write this out explicitly. Could I have a pointer towards a function or simpler way of showing that the boundary allows a transformation of this kind?