Construct your collection greedily by at each step choosing a circle of largest radius that doesn't intersect an already chosen circle. Let $T$ be the collection thus formed, and let $T'$ be the collection formed by taking each circle in $T$ and replacing it by a circle of $3$ times the radius with the same center. Any circle in our original collection intersects some circle in $T$ with radius at least as large as itself. Tripling the size of that intersecting circle causes it to contain the smaller circle. In other words, $T'$ contains $U$. Since $T'$ has area at most $9$ times the area of $T$, $T$ has area at least $1/9$ the area of $U$.