Edit 1: Changed from monotonically increasing sequence to a strictly increasing sequence.
Edit 2: Let $\{a_n\}$ denote a sequence such that for all $n\geq 2$, $a_n < a_{n+1}$ and $a_n \leq a_1$.
Is it possible to have a sequence that is strictly increasing but whose first term is the maximum?
I think not.
Let $\{a_n\}$ be a sequence such that for all $n\geq 2$ we have $$a_n < a_{n+1}$$ Suppose that $a_n \leq a_1$ for all $n\geq 2$. Then $\{a_n\}$ is an infinite increasing sequence that has a maximum, so $\{a_n\}$ is finite, which is a contradiction.
As Kavi Rama Murphy pointed out in his comment, $a_n = 0$ satisfies the criterion you described. However,it is true that there exists no strictly increasing sequence with more than a single element for which $a_0 = \text{max}\{a_n\}$ is the maximum. In particular, because $a$ is strictly increasing, $a_0 < a_1$, which implies that $\text{max}\{a_n\}_{n \leq 1} = a_1 > a_0$. Further, since $\{a_n\}_{n \leq 1} \subset \{a_n\}$, we have $\text{max}\{a_n\} \geq \text{max}\{a_n\}_{n \leq 1} = a_1 > a_0$.