Possible condition on locally Euclidean subset of Euclidean space to be embedded submanifold II Given a locally Euclidean (locally homeomorphic to some Euclidean space) subset $X\subset\mathbb R^n$ and $p\in X$, let $\widetilde{\mathrm{T}}_pX$ denote the tangent set of $X$ at $p$, namely the set of derivatives of differentiable curves in $X$ based at $p$.
Suppose for all $p\in X$ we have that $\widetilde{\mathrm{T}}_pX$ is a linear subspace of $\mathbb R^n$ of dimension $\dim_pX$. Does it follow that $X\subset\mathbb R^n$ is an embbeded differentiable submanifold?
Added. Here's a thought. Perhaps we can locally construct an exponential map which is a diffeomorphism between a neighborhood of the tangent plane and a neighborhood of $p$ in $X$. Using this diffeomorphism we can move between a differentiable structure on $X$ and the differentiable functions defined on the aforementioned neighborhood of the tangent plane.

 A: Construct a function $f:\mathbb R^2\to\mathbb R$ by putting a smooth bump of height $2^{-n}$ supported in a ball of radius $2^{-3n}$ around $(0,2^{-n}),$ for each positive integer $n.$ This function is smooth except at $(0,0).$ Let $X$ be the graph $\{(x,y,f(x,y))\}.$ The projection onto the first two-co-ordinates is a homeomorphism onto $\mathbb R^2,$ and restricts to a diffeo from $X\setminus\{(0,0,0)\}$ to $\mathbb R^2\setminus\{(0,0)\}$ so the tangent sets are good except possibly at $(0,0,0).$ For $(0,0,0),$ for each desired tangent vector $(x,y,0)$ not parallel to $(0,1,0)$ we can take a straight line $(tx,ty,0)$ ($t$ small) as the tangent curve. And for $(0,1,0)$ we can take $(t^2,t,0).$
A: This MO answer also implies a negative answer to my question. The second subset constructed there does not admit a differentiable structure making its inclusion into a differentiable embedding.
Here is a rough drawing and here is a rough description of why this example satisfies the condition in question.
