 Solution

Every strictly increasing sequence is in ${\mathbb{N}}^{\mathbb{N}}$, whose cardinality is ${\aleph }_0^{{\aleph }_0}=\mathfrak{c}$. This gives the upper bound.

For each binary sequence $ϵ=({ϵ}_n{)}_{n\in \mathbb{N}}$, defineThen $a_{n+1}-a_n=2+{ϵ}_{n+1}-{ϵ}_n\ge 1$, so the sequence is strictly increasing. Distinct binary sequences give distinct increasing sequences. This injects a set of size $2^{{\aleph }_0}$ into the desired family, giving the matching lower bound. HenceOne may also identify an increasing sequence with its infinite range subset of $\mathbb{N}$; the explicit binary coding establishes the lower bound without discarding finite subsets by cardinal subtraction.