If we use "L" to denote "Late" and "NL" to denote "Not Late", there are four things that can happen over the two days:
(L, L)
(L, NL)
(NL, L)
(NL, NL)
The sum of the probabilities of all of the above is 1.
The probability of being late on at least one of the two days is P(L, L) + P(L, NL) + P(NL, L). As P(L, NL) is the same as P(NL, L), this can be re-written as P(L, L) + 2P(L, NL). Or, as they've done in the mark scheme, the probability of being late on at least one of the two days is 1 - P(NL, NL).
So why does your approach give the wrong answer? You've correctly worked out the value of P(L), but you need to consider that when you look over the two days, P(L) = P(L, L) + P(L, NL). If in doubt about that, draw a new tree with L and NL as its branches, with two stages. From that it follows that 2P(L) = 2P(L,L) + 2P(L, NL). The upshot is that you've ended up double-counting the (L, L) case so your answer is a little on the high side.