I came up with something that seems to be proving the assertion. Though, I would not say that I am sure that everything in it is correct. Could someone please read through and comment on it?
Here it goes: First I define a concept that I use in the proof. I have no idea if what I define is consistent with the usual definition of cycle structure, but it works for the purpouse of this proof.
Definition: If \sigma \in S{n} we know that we can write it as a product of disjoint cycles. We characterize \sigma in the following way;
If \sigma is a product of \tau_{k_{1}}, \ldots \tau_{k_{r}}, where the \tau_{k_{i}} are disjoint cycles of length k_{i}, we say that the cycle structure of \sigma is \sum{k_{i}}, denoted \mathcal{S}(\sigma) = \sum{k_{i}}.
Proof: Assume H \leq A_{5}, |H| = 20. If \sigma \in H then \mathcal{S}(\sigma) \in \{\phi,\: 2+2,\: 5\}, as if the structure would have been 2,4,3+2 it would have been an odd permutation. As the order of a 3-cycle is 3, we get by Lagrange's Theorem that no 3-cycle can be an element of a group of order 20. Thus we know that any element \sigma \in H must be either a product of two disjoint transpositions, a 5-cycle or a product of these. We now examine the subgroups of A_{5} having only elements of said type.
Let \tau and \sigma be of structure 2+2 and 5, respectively. Consider the subgroup K = \langle(\tau, \sigma)\rangle. As there exists 5 distinct powers of \sigma and 2 distinct powers of \tau (conclusion drawn by the orders of the subgroups generated by one of the elements), we know that |K| \geq 5+2-1, as the identity is counted twice. Also, as the \tau\sigma^{k} are distinct for k = 1,2,3,4,5, and only for k = 5 we have \tau\sigma^{k} = \tau, we get |K| \geq 10. If \tau conjugates \sigma to a power of itself, equality holds. Even if one would not guess that this tells us anything, it actually reduces the problem quite alot.
If we are to arrive at a subgroup of order 20 we must adjoin some element of the appropriate type to the generating set. However, if we adjoin a 5-cycle \sigma', we have that, if K' = \langle(\tau, \sigma', \sigma)\rangle, |K'| \geq 40, so it will generate the whole of A_{5}. If we adjoin another double-transposition \tau' perhaps? Well, if K'' = \langle(\tau, \tau', \sigma)\rangle, for every \rho \in K we obtain \tau'\rho \in K'' \backslash K, and they are by the cancellation laws with respect to \tau' all distinct. Thus |K''| \geq 20. Can we have equality? Well, no, as \tau and \sigma does not commute. This is due to the fact that, if they would commute, so \tau\sigma = \sigma\tau, then \tau\sigma\tau = \tau^{2}\sigma = \sigma. But if \sigma = (a_1, a_2, a_3, a_4, a_5) and \tau = (a_1, a_k)\tau_x where \tau_x is a transposition (we can always modify \sigma to begin with the same letter as \tau), then \tau\sigma\tau moves a_1 to a_k, further to a_{k+1} and we would be forced to require \tau to move a_k+1 back to a_{2}. Thus \tau = (a_1, a_k)(a_k + 1, a_2), so \tau\sigma\tau moves a_2 to a_{k+1}, and then to a_{k+2}. As \sigma moves a_2 to a_3, we would need k+2 = 3, so k=1. But then \tau would certainly not be a double transposition, which is a contradiction.
Thus we have |K''| > 20. As every step in the construction was necessary to even keep the possibility of K'' = H, the supposed subgroup of order 20 of A_5, we can be sure that no such subgroup exists.