partial solution

Something like this should work. We are given $r_1, r_2, r_3 > 0$ with $r_1^2+r_2^2+r_3^2<1$. Let $s$ be such that $r_1^s+r_2^s+r_3^s = 1$; this $s$ is the similarity dimension of our IFS. Then $0 < s < 2$.

We will choose three fixed point $p_1, p_2, p_3$ in the plane. Our three maps will be $f_1(x) = r_1 (x - p_1)+p_1$, $f_2(x) = r_2 (x - p_2)+p_2$, $f_3(x) = r_3 (x - p_3)+p_3$. So $f_i$ has fixed point $p_i$ and contraction ratio $r_i$. Let $K$ be the attractor of the IFS $(f_1,f_2,f_3)$. That is, $K$ is a nonempty compact set with $K = f_1(K) \cup f_2(K) \cup f_3(K)$.

By a theorem of K. Falconer, for almost all choices of the three points $p_1, p_2, p_3$, the Hausdorff dimension of $K$ is $s$.

In fact, it should be true (since $s < 2$) that for almost all choices of $p_1,p_2,p_3$, the images $f_1(K), f_2(K),f_3(K)$ are pairwise disjoint. IF that happens, then an $\epsilon$-neighborhood of $K$ will work as the open set $O$ for small enough $\epsilon$.

Even if the images $f_1(K), f_2(K),f_3(K)$ are not pairwise disjoint, I seem to recall a theorem to the effect that if the Hausdorff dimension coincides with the similarity dimension, then the open set condition must hold (which is the existence of open set $O$ requested in the theorem).

partial solution

Something like this should work. We are given $r_1, r_2, r_3 > 0$ with $r_1^2+r_2^2+r_3^2<1$. Let $s$ be such that $r_1^s+r_2^s+r_3^s = 1$; this $s$ is the similarity dimension of our IFS. Then $0 < s < 2$.

We will choose three points $p_1, p_2, p_3$ in the plane. Our three maps will be $f_1(x) = r_1 (x - p_1)+p_1$, $f_2(x) = r_2 (x - p_2)+p_2$, $f_3(x) = r_3 (x - p_3)+p_3$. So $f_i$ has fixed point $p_i$ and contraction ratio $r_i$. Let $K$ be the attractor of the IFS $(f_1,f_2,f_3)$. That is, $K$ is a nonempty compact set with $K = f_1(K) \cup f_2(K) \cup f_3(K)$.

By a theorem of K. Falconer, for almost all choices of the three points $p_1, p_2, p_3$, the Hausdorff dimension of $K$ is $s$.

In fact, it should be true (since $s < 2$) that for almost all choices of $p_1,p_2,p_3$, the images $f_1(K), f_2(K),f_3(K)$ are pairwise disjoint. IF that happens, then an $\epsilon$-neighborhood of $K$ will work as the open set $O$ for small enough $\epsilon$.

Even if the images $f_1(K), f_2(K),f_3(K)$ are not pairwise disjoint, I seem to recall a theorem to the effect that if the Hausdorff dimension coincides with the similarity dimension, then the open set condition must hold (which is the existence of open set $O$ requested in the theorem).