= 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> $\epsilon=(\epsilon_n)_{n\in\mathbb N}$, define
$$
a_n=2n+\epsilon_n.
$$
Then $a_{n+1}-a_n=2+\epsilon_{n+1}-\epsilon_n\ge1$, 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>. Hence
$$
\boxed{\text{cardinality}=\mathfrak c.}
$$
One 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.
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