A topological surface can be pretty strange (consider, for instance, covers of $S^{2}K$, where $K$ is a Cantor set.) Can every orientable topological surface be topologically embedded in $\mathbb{R}^{3}$?

$\begingroup$ Similar question was recently discussed here, but I don't now how to find it. Are you asking about topological surface (and topological embedding)? Or a Riemann surface (conformal embedding) or a surface with some metric, or what? I believe that a topological surface, and even a Riemann surface can be so embedded. $\endgroup$ – Alexandre Eremenko Nov 16 '12 at 1:50

1$\begingroup$ David  if you are not requiring the embedding to be proper then this should follow from the classification theorem, see e.g. jstor.org/discover/10.2307/… My guess is that starting from this one can construct a proper embedding too. $\endgroup$ – algori Nov 16 '12 at 1:58
Expanding slightly on my comment above: here is how one can get the embedding theorem from the classification theorem.
The classification theorem for noncompact surfaces (theorem 3 in http://www.ams.org/journals/tran/196310602/S00029947196301431860/S00029947196301431860.pdf; by the way, the comment links to a version that tries to charge one $30 unless one's reading this from a university account; sorry about that; this has been fixed now) states that
Every surface is homeomorphic to a surface formed from a sphere $\Sigma$ by first removing a closed totally disconnected set $X$ from $\Sigma$, then removing the interiors of a finite or infinite sequence $D_1,D_2,\ldots$ of nonoverlapping closed discs in $\Sigma  X$, and finally suitably identifying the boundaries of these discs in pairs. (It may be necessary to identify the boundary of one disc with itself to produce an odd "cross cap.'')[...]
To get a proper embedding in $\mathbb{R}^3$ first note that one may assume that $X$ is nonempty; otherwise the surface will be compact (since we are throwing away something open and then identifying something), in which case everything is clear. Let $f:\Sigma\to [0,\infty)$ be a smooth function such that $f^{1}(0)=X$.
Take an $x_0\in X$ and identify $\Sigma\setminus \{x_0\}=S^2\setminus \{x_0\}$ with $\mathbb{R}^2\subset\mathbb{R}^3$. First we properly embed $\SigmaX=\mathbb{R}^2\setminus X$ in $\mathbb{R}^3$ as the graph of the function $\frac{1}{f}$.
Let $D_i,D_j$ be two disks whose boundaries are to be identified. There is an $\varepsilon(D_i,D_j)>0$ such that one can join $D_i$ and $D_j$ with a curve $\gamma(D_i,D_j)\subset \Sigma$ that misses $f^{1}([0,\varepsilon(D_i,D_j)))$. Now delete the portion of the graph over the interiors of $D_i$ and $D_j$, attach vertical tubes ("chimneys") that reach at least as high as $1/\varepsilon(D_i,D_j)$ to the resulting boundaries, and then connect the tops of the chimneys with a horizontal tube along $\gamma(D_i,D_j)$.
If $D_k,D_l$ are two other disks that we must throw away and then glue the boundaries we proceed in a similar way but this time we may have to make the chimneys higher so that they miss the first tube. And so on. It is not hard to see that each ball centered at the origin intersects only finitely many tubes.
Here is a reference: MR0304649 (46 #3781) Rüedy, Reto A. Embeddings of open Riemann surfaces. Comment. Math. Helv. 46 (1971), 214–225.
It talks about Riemann surfaces, but every orientable topological surface (with a countable basis of topology) has a structure of a Riemann surface.
I suppose there is no question about embedding of closed orientable surfaces.