What is the smallest graph that contains all non-isomorphic 4-node and 5-node connected graphs as induced subgraphs?

Surprisingly, the answer is 7 vertices for all 4-vertex subgraphs and 9 vertices for all 5-vertex subgraphs. I checked all graphs with 8 vertices and this was the first one (and only one so far) with 9 vertices that I found:

This is the listing of each of the 21 different subgraphs with 5 vertices:

This is a graph which contains all the graphs with 4 vertices:

That is, $K_4 : \{a,b,c,d\}$, $K_4-edge : \{ a,b,c,e \}$, $C_4 : \{a,b,e,f \}$, $P_4 : \{ c,d,e,f \}$, $K_{1,3} : \{b,e,f,g\}$ and the final one $\{b,c,d,f\}$.

And this is a graph with 10 vertices which contains all connected graphs with 5 vertices:

This is all of the different subgraphs of the latter graph:

I'll continue the search for a 9 vertex graph overnight, but I'm now doubting it can exist. I can also update my answer with the Maple code if anyone is interested.

Even with the exponential increase in number of small subgraphs, there are so many graphs that I'd like to believe that there would be a graph with all subgraphs on $k$ vertices which isn't exponential in $k$. (but I would be wrong!)

The size of a graph that contains all $k$-vertex subgraphs as induced subgraphs is exponential in $k$.

There are more than $2^{\binom{k}2}/k!$ graphs on $n$ vertices. So if $X$ is a graph on $n$ vertices that contains all $k$ vertex subgraphs as induced subgraphs then $$ \binom{n}{k} \ge 2^{\binom{k}2}/k! $$ and as $$ \binom{n}{k} \le \frac{n^k}{k!} $$ we have $$ n^k \ge 2^{\binom{k}2}. $$ So $$ k\, \log_2(n) \ge k(k-1)/2 $$ and therefore $n \ge 2^{(k-1)/2}$.

Let $p$ be a prime congruent to 1 mod 4. The vertices of the Paley graph are the integers mod $p$; two integers are adjacent if their difference is a square mod $p$. The Paley graph on $p$ vertices contains all graphs on $k$ vertices as induced subgraphs, where $k\approx c\,\log_2(p)$.