Notice that given a division of the array into strictly increasing subsequences, it is always possible to make the endpoints of these subsequences decreasing (not strictly).
Why? well, if we have a subsequence $s_i$ whose endpoint is smaller than the endpoint of any of the following subsequences $s_j$ (with $i < j)$, we can assign the endpoint of $s_j$ to $s_i$. By applying this reasoning as many times as desired we can conclude that indeed, given a partition of the initial array into strictly increasing subsequences, it is possible to obtain a new partition with at most the same amount of subsequences, and such that the endpoints are decreasing.
Is it clear that if $d$ is the length of the longest decreasing subsequence, we will not be able to obtain a partition of the array in less than $d$ strictly increasing subsequences.
Furthermore, using what we just proved, we can see that we can always make a partition with at most $d$ increasing subsequences. Given an initial partition, and applying the aforementioned process, we can always obtain a partition of the array such that the endpoints are decreasing. However, as the longest decreasing subsequence has length $d$, it is not possible that this partition has more than $d$ subsequences, since the endpoints would form a decreasing subsequence of length more than $d$.
Hence the solution is exactly the length of longest decreasing subsequence.