No need for Bayes.
Observe that by the definition of conditional probability:
$ P(B|A)P(A) = P(A \text{ and } B) $
and that
$ P(A \text{ and }B) + P(A' \text{ and } B) = P((A \text{ or }A') \text{ and } B) = P(B) $
So we have
$ P(B) = P(B|A) P(A) + P(B|A') P(A') \tag{%}$
If $P(B|A)> P(B)$ and $P(B|A') > P(B)$, equation (%) implies that
$ P(B) > P(B) P(A) + P(B) P(A') = P(B)\left[ P(A) + P(A')\right] = P(B) $
which is a contradiction.
Intuitively this is the statement that "the weighted average of two numbers must lie between the two numbers". $P(B)$, the probability of $B$ happening over the whole sample space, is the weighted average of $P(B|A)$, the probability of $B$ happening over the sample space for which $A$ happens, and $P(B|A')$, the probability of $B$ happening over the remaining part of the sample space. Hence we must have that if $P(B|A) > P(B)$, the left over bit satisfies $P(B) > P(B|A')$.