Why can all surfaces with boundary be realized in $\mathbb{R}^3$? I'm having trouble comprehending an informal proof of the fact that all compact surfaces with boundary can be realized in $\mathbb{R}^3$. I'm trying to find a proof of it on the internet, but I can't find it anywhere. Could someone please give a reference or give a proof of it, starting from the classification of compact surfaces (with or without boundary) ?
The proof I'm trying to understand starts from a disk $D$ and glues rectangular strips to change the Euler characteristic and the boundary number (= the number of holes made in $D$), but I don't understand how the gluing happens.
Any help is as always appreciated.

From the classification theorem, a compact surface with boundary is either $mT+nD$ or $mP+nD$, where $T$ is a torus, $P$ a projective plane, $D$ a disk, $+$ is "connected sum", $m\ge0$, and $n\ge1$. 
Now you can certainly embed $T$ in 3-space, and the connected sum of any number of copies of $T$, and then you can cut some holes to get $mT+nD$, so the orientable surfaces give no problem. 
$P+D$ is a Möbius band, which you've surely seen embedded in 3-space. 
$2P+D$ is a Klein bottle with a hole in it. The usual way of trying to sketch a Klein bottle suffers from the defect that there's a place where the surface has to pass through itself; if you carve a hole there, the surface can pass through the hole. Alternatively, think of a square with opposite edges identified in a Klein bottle way, but with a hole in the middle of the square. Identify one pair of edges to make a cylinder with a hole in the wall, then identify the other pair of edges (which at this point are circles) by pulling one through the hole to meet the other. 
Finally, $3P$ is homeomorphic to $T+P$, and you can use this to write $mT+nD$ as some number of tori plus either $P+D$ or $2P+D$, plus some more copies of $D$, so you always get something that lives in 3-space. 
