An orientable surface that cannot be embedded into $\Bbb R^3$? I previously asked this question on MSE, without success.

By Whitney's embedding theorem, every 2-dimensional manifold (aka. a surface) can be embedded into $\Bbb R^4$.
Now, Wikipedia states in this paragraph that we can even embedd into $\Bbb R^3$ if the surface is


*

*compact and orientable, or

*compact and with non-empty boundary.


In the second bullet point, it is clear that I cannot drop either of the conditions. 
It is not clear to me why I can't drop "compact" in the first bullet point.

Question: Is there an orientable but non-compact surface that does not embedd into $\Bbb R^3$?

 A: It seems to me that every orientable surface is indeed embeddable in $\mathbb{R}^3$. By Ian Richards' classification theorem (https://www.ams.org/journals/tran/1963-106-02/S0002-9947-1963-0143186-0/S0002-9947-1963-0143186-0.pdf), we know that a non-compact orientable surface is determine by the pair $Y\subset X$ where $X$ is its space of ends (homeomorphic to a compact subset of the Cantor space) and $Y$ is the closed subset of ends with genus.
When genus is finite, i.e. $Y=\varnothing$, the embedding is easily realised as a genus-$g$ surface with a copy of $X$ removed. When genus is infinite, one simply starts with a sphere with a copy of $X$ removed, and adds smaller and smaller handles accumulating to all points of $Y$ (but not to points of $X\setminus Y$). 
To ensure this can be done, observe that $Y$ has a countable dense subset. By taking a countable basis of neighborhoods of each of them, we get a countable family $(U_n)_{n\in\mathbb{N}}$ of open subsets of $\mathbb{S}^2\setminus X$ and we want to put one handle in each of them, but no family of handles should approach any point outside $Y$. We put a handle in $U_1$ (i.e. we remove two discs from $U_1$ and glue a handle to them), then inductively add one in the smallest $U_n$ not containing both gluing circle of any previous handle.
