It is not possible. The proof is by contradiction. Suppose that such a covering family $\mathcal{F}$ exists. Let $D(P, \rho)$ denote the disc with center $P$ and radius $\rho$. Start with an arbitrary disc $D(O, r)$ that does not overlap any member of $\mathcal{F}$. Then $D(O, r)$ covers no grid point. Take the disc $D(O, r)$ to be maximal in the sense that any further enlargement would cause it to violate the non-overlap condition. Then $D(O, r)$ is tangent to at least three discs in $\mathcal{F}$. Observe that there must be two of the three tangent discs, say $D(A, a)$ and $D(B, b)$, such that $\angle AOB \le 120^\circ$. By the Law of Cosines applied to triangle $ABO$,
$$(a+b)^2 \le (a+r)^2 + (b+r)^2 + (a+r)(b+r),$$
which yields
$$ab \le 3(a+b)r + 3r^2, \quad \text{and thus}\quad 12r^2 \ge (a-3r)(b-3r).$$
Note that $r < 1/\sqrt{2}$ because $D(O, r)$ covers no grid point, and $(a-3r)(b-3r) \ge (5-3r)^2$ because each disc in $\mathcal{F}$ has radius at least $5$. Hence $2\sqrt{3}r \ge (5-3r)$, which gives $5 \le (3+2\sqrt{3})r < (3+2\sqrt{3})/\sqrt{2}$ and thus $5\sqrt{2} < 3+2\sqrt{3}$. Squaring both sides of this inequality yields $50 < 21+12\sqrt{3} < 21+12 \cdot 2 = 45$. This contradiction completes the proof.