Sometimes a curve, or closed curve, or simple closed curve, is assumed to be piecewise smooth; even if the book doesn't say so. The jordan curve theorem assumes the jordan curve, or simple closed curve, is piecewise smooth. If you draw such a curve in the plane, it separates the plane into two pieces, the inside and the outside. This is obvious if you draw a few examples and think about it for a while, but it is notoriously difficult to prove. Even with modern mathematics at my command, I am going to gloss over more than a few details.
If the curve is not piecewise smooth, or injective, there may be no inside at all. The curve could completely fill a region in the plane. This is called a space filling curve, and an example will be given in the next section. Since this is somewhat pathological, let's assume c is injective and piecewise smooth.
Let p be a point in the middle of one of the smooth segments of c. Thus c passes through p, and moves in (more or less) one direction. The curve can't suddenly double back and touch p again, because it is practically a straight line at this point. If no other part of c touches p, then the rest of c can be bounded away from p, as described in an earlier section. This is certainly the case when c is injective. So - select points u and v, on opposite sides of c, such that the segment uv passes through p, and is perpendicular to p, and u and v are very close to p - closer than any other part of the curve. In other words, the segment uv crosses c at p, and that is its only contact with c.
Find a polygonal approximation, such that the homotopy does not touch u or v. The curve moves only slightly, if at all. It is still (nearly) perpendicular to uv, and the intersection is (nearly) the midpoint of uv.
Now apply ray intersection. Let the ray start at u and pass through v. The ray starting at u has one more crossing than the colinear ray starting at v. Thus the winding number of u is different (by ±1) than the winding number of v. The regions on opposite sides of c represent different path components. They cannot both be "outside", hence there is at least one component "inside".
The trick is to enclose c in a tube. Let u be a point far away, and draw a ray towards c, with v being the first point of intersection. If there is no "first" point of intersection, then the implied cluster point, not on c, proves the complement of c is not an open set. This is a contradiction, hence there is such a point v. Pull back from v, towards u, until you hit the wall of the tube. This is path connected with u, hence it is the "outside wall".
A line passing through v, perpendicular to c, strikes the outside wall, and the "other" wall. The former has a winding number of 0, since it is outside. The latter has a winding number of 1, since we have crossed c once to get there. This can't be an outside wall, so call it an "inside wall".
Travel along the outside wall, following the twists and turns of c, until you get back to the same cross section of the tube. A path cannot connect two points with different winding numbers, so you must be back to start, back to the original point in the outside wall. In other words, the outside wall is a jordan curve that parallels c. Similarly, the inside wall is a jordan curve that parallels c, lying entirely inside c.
Let x be a point that is inside c, but not inside the tube. Draw a ray from x in any direction. The ray must intersect c, else x has a winding number of 0, and lies outside of c. Again, there is a first intersection, and this is preceded by a wall. This cannot be an outside wall, since x is inside. Thus x is path connected to the inside wall. This holds for every x inside c, and outside of the tube. This region is path connected.
Finally, the inner wall is path connected to every x inside c, inside the tube. Embed c in a smaller tube, i.e. choose a smaller radius r, such that a perpendicular drawn from x to c cuts through the smaller tube. In other words, a regular cylinder, not at one of the corners, passes beneath x. Any slice of the larger tube connects the two inner walls, and x is also path connected to the inner wall of the smaller tube. That joins x to the inner wall of the larger tube. Everything inside c is path connected, and that completes the proof.
In summary, a piecewise smooth jordan curve (with finitely many pieces) cuts the plane into two path connected components, the inside and the outside.