Note that this condition is sufficient, but not necessary. Every metric space is normal, but may not be second countable, as shown by the discrete topology on an uncountable set. Each isolated point is open, and is the union of base sets, hence there is a base open set for each point, which contradicts second countable.
Let S be a space that is normal and second countable. Enumerate the base sets and consider them one at a time. Let O be a base open set and let C be its complement. If O is the empty set, or all of S, skip it; we don't need it.
Let U be a base open set in O, where U closure is also in O. Let D be the closure of U. Since C and D are disjoint closed sets, and S is normal, use urysohn's lemma to build a continuous function fO,U(S) that is 0 on C and 1 on D.
Build such a function fO,U for every nested pair of base sets O and U in S. This is a countable collection of functions, and they can be enumerated as fj, as j runs from 1 to infinity. Each function is a continuous map from S into [0,1]. The composite function f maps S into the product space P, a hypercube with countable dimensions. Give P the weak product topology. Since each component function fj is continuous, the composite function f is continuous. We have built a continuous map from S into P.
Consider to points x and y in S. Separate x and y in disjoint open sets, and they may as well be base open sets. Let x be contained in O, whence y is in C, the complement of O. Since x and C are closed, separate them in disjoint open sets. Let U be a base open set containing x. Let D be the closure of U. Note that D misses the open set containing C, and C itself. Once again C and D are disjoint closed sets. There is some fj, derived from the open sets O and U, and fj(y) = 0 and fj(x) = 1.
For any two points x and y, some function fj in the list maps them to different values in [0,1]. Therefore the composite function f is 1-1.
We know f is continuous; let's show it is bicontinuous. It is enough to show that the image of O, for any base set O, is the union of open sets in P.
Let C be the complement of O. Select any x in O and build U and D as we did before. Let fj map C to 0 and D to 1. Consider the open set in P produced by restricting the jth coordinate to values greater than one half. The other coordinates are unconstrained. Intersect this open set with f(S) to get an open set W in the image of S. Let V be the preimage of W.
The point z lies in V iff fj(z) exceeds one half. Clearly z cannot lie in C, hence V is wholly contained in O. The image of O under f covers all of W. Since fj(x) = 1, x is in V. We have accounted for the image of x.
Do this for all x in O, and O is the union of open sets Vx, each having an open image Wx in f(S). Put this all together and the image of O is an open set. Therefore f is bicontinuous, and a homeomorphism.
For all practical purposes, S is a subspace of the space P.
Since P is the countable product of metric spaces it is metrizable, as demonstrated in the previous section. This metric gives S its topology. Therefore S is metrizable.
Let S be a set, and f a collection of functions from S into [0,1]. If f has size n then f maps S into the hypercube [0,1]n. Of course n could be infinite, perhaps uncountable.
The image of S is a space inside the hypercube, which inherits the product topology.
The composite function f, from S into the hypercube, is described by the component functions f1 f2 f3 etc, where fj maps S to the jth coordinate of the hypercube. If you prefer, fj is f composed with the jth projection.
In general, the topology induced by a function such as f is the weakest topology needed to make f continuous. The preimage of each open set in the product space P must be an open set in S. In particular, the preimage of each base open set in P must be an open set. Let's see what this means for our hypercube.
A base set restricts finitely many coordinates to open intervals, and leaves the rest unconstrained. A point x in S is in the preimage if fj(x) lies in the designated open interval for the jth coordinate, for those coordinates that are constrained.
We can pull this back a bit. Imagine that the open intervals have centers cj, and a common radius r. The preimage of such a set must be open, and if these preimages are all open, the preimage of an open set, with arbitrary intervals, is also open.
Let's see if this acts as a base for a topology on S. Let x be in two preimages at once. Let j loop over coordinates that are constrained by either the first or second image. This is still a finite set.
Let cj = fj(x). This sits inside an open interval, or perhaps two open intervals simultaneously.
Choose rj small so that cjħrj does not break out of either interval. Let r be less than any of the values rj.
Consider the preimage of the open set with centers cj and radius r. This contains x, and is contained in the intersection of our two preimages. The intersection is covered by base open sets, giving a base for the functional topology of S.
All of the above remains valid for a product of arbitrary metric spaces. I used the space [0,1] for convenience, but any metric space will do. Our open intervals become open balls, and everything else is the same.
Assume every pair of points in S has a function fj, that maps them to different images in the jth metric space. Thus the function f(S) is 1-1 onto its image. Base open sets in S are, by definition, preimages of base open sets in P. Taking unions, open sets in S map to open sets in P. Thus f is bicontinuous, and a homeomorphism.
This isn't too surprising. If f is 1-1 then S really is a subset of P. Let f(S) = T, the image space in P. The weakest topology that makes f continuous has to preserve the open sets of T, as determined by the topology of P. And there's no reason to have any more open sets, so S and T are homeomorphic.