Seven is minimal, and shown below. The centers of the six outer unit circles are $\sqrt3$ away from the center of the circle of radius $2$, and they form a regular hexagon. It is not hard to check that this works. [asy][asy] size(4cm); pair O=(0,0); pair A=sqrt(3)*dir(0); pair B=sqrt(3)*dir(60); pair C=sqrt(3)*dir(120); pair D=sqrt(3)*dir(180); pair E=sqrt(3)*dir(240); pair F=sqrt(3)*dir(300); draw(circle(O,2)); draw(circle(O,1)); draw(circle(A,1)); draw(circle(B,1)); draw(circle(C,1)); draw(circle(D,1)); draw(circle(E,1)); draw(circle(F,1)); [/asy][/asy] Proof that six unit circles fail: Let the circle of radius $2$ be $\Gamma$ and have center $O$, and let $\omega$ be a unit circle. Assume $\Gamma$ and $\omega$ intersect at $A$ and $B$. Since $\overline{AB}$ is a chord of $\omega$, the maximum possible length $AB$ is $2$, so the maximum possible measure of $\angle AOB$ is $60^\circ$. From the above, six unit circles can cover at most $360^\circ$ of $\Gamma$'s circumference, so equality must hold. In other words, all unit circles intersect $\Gamma$ at two points forming an arc of measure $60^\circ$, and adjacent unit disks intersect on the circumference of $\Gamma$. As seen in the diagram above, six unit circles positioned in such a manner cannot cover $\Gamma$, as desired. $\square$