Characterization of non-separating submanifolds According to the book I'm reading (Introduction to $3$-manifolds, by Jennifer Schultens), 

A submanifold $A$ of a connected manifold $X$ is separating
  if $X \setminus A$ has at least two components; otherwise it is non-separating.

The remark following this definition reads:

Connected manifolds are path connected. It follows that a submanifold $A$ of a connected manifold $X$ is non-separating if and only if there is a simple closed curve in $X$ that intersects $A$ transversely in a single point.

I was able to prove that if $A$ is non-separating then such a curve exists. Could you give me any hints on how to prove the converse?
 A: Intersection numbers of submanifolds are well-defined in homology (that is, if you're taking the intersection number of $S$ and $L$, replacing $S$ with the homologous $S'$ gives the same intersection number). A separating submanifold is null-homologous, so the intersection number with any curve is automatically zero.
In more elementary language: Let $M$ be one side of the separating submanifold $S$. Suppose $L$ is a submanifold of the opposite codimension that intersects $S$ transversely. Then $L \cap M$ is also a transverse intersection of oriented manifolds, so is an oriented compact 1-manifold with oriented boundary $L \cap S$; but because the oriented number of points on the boundary of any compact 1-manifold is $0$, the intersection number coming from $L \cap S$ is zero.
A: If you proved this, you did something wrong: it is not a true statement. You need to assume that the submanifold is connected. Otherwise, take a non-separating connected submanifold and add a parallel copy (possible e.g. assuming orientability). Now there are only curves which transversly intersect this submanifold in an even amount of points. Maybe you want to show us your proof?
For an elementary proof of the other direction note that connectedness equals path-connectedness in this context. Given a curve intersecting $A$ only once in $x$, you can connect any two points in $X\setminus A$ by taking a connecting path in $X$ and do the following: use the connectedness of $A$ to connect all intersection points of this connecting path to $x$; modify the loop for every intersection point by going through the complement via the given curve intersecting $A$ once; push the new loop of $A$, giving you a connecting curve in $X\setminus A$. (note that the idea is replacing the intersections with $A$ by going through the complement).
If you know that every 1-codimensional submanifold gives you a homomorphism on loops by assigning intersection numbers, this gives you a more elegant one-liner to prove it, cf. Mike Miller's answer.
