Use the following strategy $S$: Put each card $c$ on the stack with the smallest top card $c'$ such that $c' > c$.
Consider a generic strategy $T$, and iteratively simulate $S$ and $T$.
Let $s_i = (s^i_1, s^i_2, \dots, s^i_{h_i})$ be the values on the top of the stacks after placing the $i$-th card $c_i$ according to strategy $S$, in increasing order. Define $t_i = (t^i_1, t^i,_2, \dots, t^i_{k_i})$ similarly for strategy $T$.
Claim: $\forall i=1,\dots,n$, we have: i) $h_i \le k_i$ and ii) $s^i_j \ge t^i_j$ for all $j = 1, \dots, h_i$.
The proof is by induction on $i$.
For $i=1$ we have $h_1=k_1=1$ and $s^1_1 = t^1_1 = c_1$.
For $i>1$ we have two cases.
Case 1) $S$ places $c_i$ in a new stack. Then, by hypothesis we must have $c_i > s^{i-1}_j \ge t^{i-1}_j$ for every $j = 1, \dots, h_{i-1} $. It follows that $k_i \ge h_{i-1} + 1 = h_i$, proving i). Notice that ii) is also verified since, for $j=1,\dots, h_{i-1}$, we have $c_i > s^i_j = s^{i-1}_j \ge t^{i-1}_j = t^i_j$ by induction hypothesis, and $t^i_{h_i} \le c_i = s^i_{h_i}$.
Case 2) $S$ places $c_i$ in an existing stack. Let $\ell$ be the index such that $s^{i}_\ell = c_i$.
We clearly have $k_i \ge k_{i-1} \ge h_{i-1} = h_i$, proving i).
As far as ii) is concerned, for every $j = 1, \dots, \ell-1$ we have $c_i > s^{i-1}_j \ge t^{i-1}_j$, and hence $s^i_j \ge t^i_j$.
Also, $s^i_\ell = c_i = \min\{s^{i-1}_\ell, c_j\} \ge \min\{t^{i-1}_\ell, c_j\} = t^i_\ell$.
Finally, for $j=\ell +1, \dots, h_i$, we have: $s^i_j = s^{i-1}_j \ge \max\{c_i, s^{i-1}_j\} \ge \max\{c_i, t^{i-1}_j\} \ge t^i_j$.
Instantiating the claim with $i=n$ yields $h_n \le k_n$, showing that $S$ is minimizing the number of stacks.
