Yes. Represent the problem as a graph with 7 vertices each with degree 4. We wish to find two disjoint triangles. First, a triangle must exist, as take any $a$ with neighbors $b, c, d, e$ and $b$ must have a neighbor among $c, d, e$ (or there would be at least 8 vertices). WLOG this triangle is $T: a, b, c$ and the remaining four are $S: d, e, f, g$. There are 6 edges between $T$ and $S$ (as there must be a total of 12 edges ending in $T$), and since there is are 16 edges ending in $S$, there must be 10 edges starting and ending in $S$, or 5 edges fully contained in $S$. By pigeonhole WLOG $d$ has three neighbors in $S$ (all the other neighbors). If no triangle in $S$ exists, $e, f, g$ cannot be neighbors. This results in only 3 edges fully contained in $S$, a contradiction. Thus, we can find two disjoint triangles, as desired.