Since L is finite over K it is algebraic, with a finite number of generators, where each is the root of a polynomial that splits in L/K. Let S be the set of these polynomials - a finite set. Thus L/K is the splitting field for S.
Since L/K is galois the polynomials of S are separable. These same polynomials split over E, hence L is a finite galois extension of E.
Start with M and adjoin the roots of the polynomials in S. The result contains M and L, and is L+M. The polynomials in S are still separable, hence L+M over M is normal and separable, and galois. Since S is a finite set, L+M is a finite galois extension of M.
Note that M/K need not be a finite extension; it doesn't even have to be algebraic.
Build a map from the M automorphisms of L+M onto the E automorphisms of L. How do we do this? Given an automorphism of L+M that fixes M, restrict this automorphism to L. Of course it automatically fixes E. Suppose it moves elements of L outside of L. This can't happen, because L is galois over E, hence it is stable within M. So the map is well defined. Every automorphism on L+M induces an automorphism on L.
This map is a group homomorphism. In other words, two automorphisms on L+M can be composed, then restricted to L, and the result is the composition of the two restrictions to L.
If an automorphism fixes both M and L it fixes the generators of L/K, which we adjoined to M to build L+M. In other words, the automorphism fixes L+M, and is trivial. The kernel of the group homomorphism is trivial. The galois group of L+M over M embeds in the galois group of L over E.
Surjective is the only tricky part. Start with an automorphism of L that maps u to v, where u and v are conjugates of the same irreducible polynomial p(x), having coefficients in E. We may as well let u be one of the generators of L/K, hence p is a factor of one of the polynomials in S. Therefore L contains all the roots of p(x).
Since u moves to v, it is not in E, and neither is v. In fact none of the roots of p(x) is in the base field E. Each of these roots generates an extension of E whose dimension is the degree of p. Therefore none of the roots of p lie in M.
Suppose p factors over M. Let g(x) be one of these factors. Write g as the product of monomials x-uj, for various conjugates of u. The coefficients of g lie in L. They also lie in M, hence they lie in E. Therefore p factors over E. Since p is irreducible, this is a contradiction. Thus p remains irreducible over M.
Now u and v are conjugates of the same irreducible polynomial over M. This means M(u) and M(v) are isomorphic field extensions of M. Mappint u to v induces a field isomorphism from M(u) onto M(v), which restricts to a field isomorphism from E(u) onto E(v).
If E(u) is not all of L, find another generator w, outside of E(u), that moves to y under the automorphism of L. (Of course w can move to w - that's ok.) Now w is the root of an irreducible polynomial p(x) over E(u). Remember that the roots of p all lie in L. If g is a factor of p its coefficients lie in L. If they also lie in M(u) they lie in E(u), and that is a contradiction. Once again p remains irreducible over M(u). Extend the isomorphism so that M(u)(w) maps onto M(v)(y).
Repeat the above process. Since L is a finite extension of E, this process terminates. (If you don't feel comfortable building towers of extensions and isomorphisms, select u so that L = E(u), which we can do, since a finite separable extension is simple.) The resulting isomorphism acts on M and all the generators of L, hence it is an isomorphism from L+M into itself. Since the dimension of L+M over M is finite, an embedding of L+M into itself is automatically onto, i.e. an automorphism. this automorphism fixes M, and induces the given automorphism when restricted to L. The map is surjective, and the galois groups are isomorphic.
Since the order of the galois group equals the dimension of the galois extension, we have an immediate corollary - the dimension of L+M over M equals the dimension of L over E.