Solution:

The answer is $15$.

Note that there are $14$ primes at most $\sqrt{2022}$, starting with $2$ and ending with $43$. Thus, the following partition works for $15$ sets. Let $S_1 = \{2, 4, \ldots, 2022\}$, the multiples of $2$ in $S$. Let $S_2 = \{3, 9, 15, \ldots, 2019\}$, the remaining multiples of $3$ in $S$ not in $S_1$. Let $S_3 = \{5, 25, 35, \ldots, 2015\}$, the remaining multiples of $5$, and so on and so forth, until we get to $S_{14} = \{43, 1849, 2021\}$. $S_{15}$ consists of the remaining elements, i.e., $1$ and those numbers with no prime factors at most $43$, i.e., the primes greater than $43$ but less than $2022$: $S_{15} = \{1, 47, 53, 59, \ldots, 2017\}$.

Each of $S_1, S_2, \ldots, S_{14}$ satisfies (a), while $S_{15}$ satisfies (b).

We show now that no partition in $14$ subsets is possible. Let a Type 1 subset of $S$ be a subset $S_i$ for which (a) is true and there exists an integer $d > 1$ for which $d$ divides every element of $S_i$. Let a Type 2 subset of $S$ be a subset $S_i$ for which (b) is true. Finally, let a Type 3 subset of $S$ be a subset $S_i$ for which (a) is true that is not a Type 1 subset. An example of a Type 3 subset would be a set of the form $\{pq, qr, pr\}$ where $p, q, r$ are distinct primes.

Claim: Let $p_1 = 2, p_2 = 3, p_3 = 5, \ldots$ be the sequence of prime numbers, where $p_k$ is the $k$th prime. Every optimal partition of the set $S(k) := \{1, 2, \ldots, p_k^2\}$, i.e., a partition with the least possible number of subsets, has at least $k-1$ Type 1 subsets. In particular, every optimal partition of this set has $k+1$ subsets in total. To see how this follows, we look at two cases:

- If every prime $p \leq p_k$ has a corresponding Type 1 subset containing its multiples, then a similar partitioning to the above works: Take $S_1$ to $S_k$ as Type 1 subsets for each prime, and take $S_{k+1}$ to be everything left over. $S_{k+1}$ will never be empty, as it has $1$ in it. While in fact it is known that, for example, by Bertrand's postulate there is always some prime between $p_k$ and $p_k^2$ so $S_{k+1}$ has at least two elements, there is no need to go this far—if there were no other primes you could just move $2$ from $S_1$ into $S_{k+1}$, and if $k > 1$ then $S_1$ will still have at least three elements remaining. And if $k = 1$, there is no need to worry about this, because $2 < 3 < 2^2$.

- On the other hand, if $p \leq p_k$ has no corresponding Type 1 subset, then $p$ and $p^2$ will not be contained in a Type 1 set. Neither can $p$ nor $p^2$ be contained in a Type 3 set. If $\operatorname{gcd}(p, x) > 1$ for all $x$ in the same set as $p$, then $\operatorname{gcd}(p, x) = p$, which implies that $p$ is in a Type 1 set with $d = p$. Similarly, if $\operatorname{gcd}(p^2, x) > 1$ for all $x$ in the same set as $p^2$, then $p \mid \operatorname{gcd}(p^2, x)$ for all $x$, and so $p^2$ is in a Type 1 set with $d = p$ as well. Hence $p$ and $p^2$ must in fact be in Type 2 sets, and they cannot be in the same Type 2 set (as they share a common factor of $p > 1$); this means that the optimal partition has at least $k+1$ subsets in total. A possible equality scenario for example is the sets $S_1 = \{1, 2, 3, 5, \ldots, p_k\}$, $S_2 = \{4, 9, 25, \ldots, p_k^2\}$, and $S_3$ to $S_{k+1}$ Type 1 sets taking all remaining multiples of $2, 3, 5, \ldots, p_{k-1}$. This works, as $p_k$ and $p_k^2$ are the only multiples of $p_k$ in $S(k)$ with no prime factor other than $p_k$ and thus cannot be classified into some other Type 1 set.

To prove our claim: We proceed by induction on $k$. Trivially, this is true for $k = 1$. Suppose now that any optimal partition of the set $S(k)$ has at least $k-1$ Type 1 subsets, and thus at least $k+1$ subsets in total. Consider now a partition of the set $S(k+1)$, and suppose that this partition would have at most $k+1$ subsets. From the above, there exist at least two primes $p, q$ with $p < q \leq p_{k+1}$ for which there are no Type 1 subsets. If $q < p_{k+1}$ we have a contradiction. Any such partition can be restricted to an optimal partition of $S(k)$ with $p < q \leq p_k$ having no corresponding Type 1 subsets. This contradicts our inductive hypothesis. On the other hand, suppose that $q = p_{k+1}$. Again restricting to $S(k)$ gives us an optimal partition of $S(k)$ with at most $k-1$ Type 1 sets; the inductive hypothesis tells us that this partition has in fact exactly $k-1$ Type 1 sets and two Type 2 sets from a previous argument establishing the consequence of the claim. However, consider now the element $pq$. This cannot belong in any Type 1 set, neither can it belong in the same Type 2 set as $p$ or $p^2$. Thus in addition to the given $k-1$ Type 1 sets and $2$ Type 2 sets, we need an extra set to contain $pq$. Thus our partition of $S(k+1)$ in fact has at least $k-1 + 2 + 1 = k+2$ subsets, and not $k+1$ subsets as we wanted. The claim is thus proved.

Returning to our original problem, since $p_{14} = 43 < \sqrt{2022}$, any partition of $S$ must restrict to a partition of $S(14)$, which we showed must have at least $15$ sets. Thus, we can do no better than $15$.