First we derive some small values and then we use two lemmata to show that $f(n)=n+2$ for $n\ge 3$. We call a collection of sets good if the condition of the problem is satisfied, hence no inclusions and distinct cardinalities. - we have $f(1)=0$ if you consider this case. - suppose $A_1,A_2\subset [1,2]$ is good . Then the $|A_k|\in \{1,2\}$ are distinct, hence one of them equals 2. Contradiction. With $\{1\},\{2,3\}$ we have $f(2)=3$. - suppose $A_1,A_2,A_3\subset [1,4]$ is good. Then the $|A_k|\in \{1,2,3\}$ are distinct, hence wlog $A_3=\{4\}$ and $A_1,A_2\subset \{1,2,3\}$ with cardinalities 2,3. Hence one of them is $\{1,2,3\}$, contradiction. With $\{1,4\},\{2,3,4\},\{5\}$ we see that $f(3)=5$. Lemma 1. Suppose $A_1,\cdots, A_n\subset [1,n+2]$ are good with cardinalities $1,\cdots,n$. Then we can construct the same for $n+2$. Proof: let $B_k:=A_k\cup \{n+3\}, B_{n+1}:=[1,n+2],B_{n+2}=\{n+4\}$. With induction base $n=2,3$ we find that $f(n)\le n+2$ for all $n\ge 2$. Lemma 2. Suppose that $n\ge 3$ and $f(n)\le n+1$. Then also $f(n-1)\le n$. Proof: suppose that $A_1,\cdots,A_n\subset [1,n+1]$ is good. The $|A_k|\in [1,n]$ are distinct hence wlog $A_n=\{n+1\}$. Then $A_1,\cdots, A_{n-1}\subset [1,n]$ are good. Now suppose that $f(n)\le n+1$ for some $n\ge 3$. Then $n>3$ and we can use Lemma 2 to obtain $f(n-1)\le n$ etc until $f(3)\le 4$. Contradiction, hence $f(n)=n+2$ for $n\ge 3$.

Above: nice solution. BTW, you can prove that $f(n)\ge n+2$ for $n\ge 3$ directly: suppose that $A_1,\cdots,A_n\subset [1,n+1]$ is good. Then since $A_i\not\subset A_j$ and $|A_i|\ne |A_j|$ for $i\ne j$ and $n\ge 3$, the sizes of $A_i$ are $1,\ldots,n$. Hence we can assume WLOG $A_n=\{n+1\}$. Then $A_1,\cdots, A_{n-1}\subset [1,n]$ are good of sizes $2,\ldots,n$. WLOG $A_{n-1}=\{1,\ldots,n\}$. But then $A_{n-2}\subset A_{n-1}$ (here we use the assumption $n\ge 3$), contradiction.

@math90: Thank you. What you write is nearly correct, we don't have inclusions in a good collection hence $A_i\subset A_j$ is not true when $i<j$. The cardinalities are distinct, however, and that's why we have $1,\cdots,n$ for the cardinalities. The rest of your argument is fine.

Thanks, corrected now.