For convenience, focus on one of the n+1 coordinates, which I will call x. Let the open set O, in projective space, be the complement of x. These are the lines that do not lie in the plane x = 0. Any of these lines may be represented by the point with x = 1. The other coordinates are unconstrained, building a bijection between O and Kn. Let's prove this is a homeomorphism.
Let U be closed in projective space, with radical homogeneous ideal H. Set x = 1 throughout and build a new ideal J that defines the corresponding algebraic set V in Kn. Every point in U∩O satisfies the polynomials in J, and conversely. The closed set U, in the subspace topology of O, becomes a closed set in Kn.
Let V be a closed set in Kn. For each generator of the ideal J that defines V, multiply each term by an appropriate power of x, so that the polynomial becomes homogeneous. These will generate a homogeneous ideal H. Setting x = 1 reproduces the polynomials of J. Thus a point is in V iff the corresponding line is in U∩O. Closed sets correspond, and O and Kn are homeomorphic.
Let's illustrate with the parabola z = y2 in R2. Clearly this is not homogeneous. Use x to build the polynomial xz = y2, which generates U. Set x = 1, and the parabola returns. When x = 0, y = 0 and z = 1. The parabola becomes the z axis, the point at infinity, the "other"focus. However, if we get carried away with x, and write x2z = xy2, U becomes the aforementioned variety, along with all the lines in the plane x = 0, i.e. an instance of P1. So there are different closed sets that map to V.
Let U1 and U2 map to V. Let H1 define U1 and H2 define U2. The intersection of U1 and U2 is defined by the union of H1 and H2, and it still maps to V. The projective closure U is defined by the union of all homogeneous ideals that produce V when x is set to 1. Find all the homogeneous polynomials that vanish on V, and let them generate the ideal H. In our notation, H = V′. Let U = H′, the set that vanishes on H. Therefore the closure of V is V′′, which is not unexpected.
The second homogenization of J includes the first homogenization, as described above, and any other homogeneous polynomials that vanish on V. Thus the second homogenization produces the projective closure of V.
Consider the twisted cubic curve in K3, parameterized by t, t2, and t3. An obvious set of generators is x2 = y and x3 = z. The first homogenization yields x2-wy and x3-w2z. (Here w is the extra variable that takes us into projective space.) Let these two polynomials generate H1, a homogeneous ideal with region U1. Of course U1 intersects O in V. Note that U1 includes, within the subspace w = x = 0, an instance of P1, as y and z attain nonzero values. However, our original ideal, spanned by x2-y and x3-z, includes a polynomial that is already homogeneous. Multiply x2+y times the first generator and subtract x times the second, giving y2 = xz. This is homogeneous, and it vanishes on V, though it has no instance of w. It is included in the second homogenization. When w = x = 0, y also has to equal 0. Our projective circle P1 is gone, not in the closure of V. Only the point 0,0,0,1 survives.
Let a homogeneous polynomial p be in the second homogenization of J. (Remember that J produces the closed set V.) In other words, p vanishes on V when w = 1. (I'm still using w as the "extra" variable.) If p does not contain w at all, then p, on its own, vanishes on V. This means p is in J, or perhaps rad(J). Either way, a power of p is in J, and it is already homogeneous. Bringing in a power of p is the same as bringing in p; we're still building the same algebraic set.
Assume p contains w. Pull out any instances of w that are common to all the terms of p. This can only make the algebraic set smaller, and it still vanishes on V. Finally set w = 1, the inverse of homogenization. The resulting polynomial, call it q, lies in J, or in rad(J). At this point we need to show that homogenization and product commute. We can homogenize q and square the result, or we can square q and homogenize. The powers of w carry along. I'll leave the details to you. Thus the second homogenization acts on the power of q that lies in J, and produces a power of p.
In summary, every homogeneous polynomial that vanishes on V is brought in by the second homogenization of J, which is the "true" homogenization of an ideal. Homogenize each element of J, span a homogeneous ideal H, and let H define a closed region U in projective space. This is the projective closure of V.
Conversely, assume J is prime. Homogenize J to get H, and suppose A*B (homogeneous) lies in H. The restriction of A times the restriction of B lies in J. One of these restrictions lies in J, and its homogenization lies in H. Therefore H is prime, and the projective closure is irreducible.
Projective closure carries affine varieties to projective varieties, and intersection with O performs the inverse operation. Thus the affine varieties embed in the projective varieties.
Conversely, suppose two projective varieties map onto the same nontrivial variety in Kn. Let U1 contain U2, where U2 is the projective closure of V. Intersect U1 with the hyperplane w = 0 to find another closed set U3. Now U1 = U2 union U3. This contradicts the fact that U1 is irreducible. Therefore affine varieties and projective varieties (not contained in w = 0) correspond one for one.
Assume you are given a subchain of projective varieties whose length is less than n+1. Select any point in the smallest variety, and let it have a nonzero w coordinate. Set w = 1 to map the chain over to affine varieties. within Kn, we can fill in the gaps. This gives a chain of length n+1 in Kn. Take the projective closure to find the corresponding chain of projective varieties, which contains the original subchain. Any chain of projective varieties can be expanded to a chain of length n+1, starting with a point and ending in the entire projective space.