Let $X \subset \mathbb{R}^n$ be a compact, connected non-empty subset. Does there exist an analogue to a tubular neighborhood for $X$, i.e. something like a smooth manifold with boundary that retracts to $X$? One can assume that $X$ has the homotopy type of a finite CW-complex.
I thought about using regular neighborhoods (which one could apply after maybe simplicially approximating $X$), but I think this delivers only a $PL$-manifold (and I am not sure if it really fits the desired properties). I also thought about considering the closure of an $\varepsilon$-neighborhood of $X$, but I do not know how I can guarantee that this is really a smooth manifold with boundary. Any help or reference for this topic is greatly appreciated!
In such general setup the answer is "no", because every retraction is a quotient map, and a quotient map preserves local connectedness. In particular the comb space is not a retract of any manifold it is embeded in, because it is not locally connected. And it is contractible, and thus of finite CW homotopy type.
The property you are talking about seems to be the same as being absolute neighborhood retract (ANR). It is well known that every locally finite CW complex is an ANR. And so in the case $X\subseteq \Bbb{R}^n$ it is enough to assume that it is just CW.