Amongst the four discs, choose one, say $\Delta_0$, containing the point $O$ where the diagonals of the quadrangle cross one another. Let $A_0$ be the center of $\Delta_0$, label the other three centers in circular order, $A_1, A_2, A_3$, so that the opposite angles $A_0OA_1$ and $A_2OA_3$ be not obtuse, and let $\Delta_i$ denote the disc centered at $A_i$.
Before proceeding, we take time out to state a simple, but quite useful lemma whose proof is postponed for the sake of clarity.

Lemma. Let $ABCD$ be a convex quadrangle, let $\Delta$ be a disc centered at $A$, and let $E$ be the point where the ray $AC$ emanating from $A$ crosses the boundary of $\Delta$. If the orthogonal projection of $B$ on the line $AC$ falls on the closed ray $EA$ emanating from $E$, then $\text{dist}(B, [ACD] \setminus \Delta) \ge BE$, where $[ACD]$ is the closure of the triangle $ACD$.

We now apply the lemma to show that $O$ is also covered by $\Delta_1$. To this end, let $B_0$ be the point where the ray $A_0O$ emanating from $A_0$ crosses the boundary of $\Delta_0$. Since $\Delta_0$ contains $O$, the latter lies on the closed segment $A_0B_0$, and since the angle $A_0OA_1$ is not obtuse, it follows that $A_1O \le A_1B_0$. On the other hand, $\Delta_1$ contains points of the closure $[A_0A_2A_3]$ of the triangle $A_0A_2A_3$ not covered by $\Delta_0$, so the radius of $\Delta_1$ is greater than or equal to $\text{dist}(A_1, [A_0A_2A_3] \setminus \Delta_0)$, which in turn is greater than or equal to $A_1B_0$ by the lemma. Consequently, $O$ is indeed covered by $\Delta_1$.

We are presently going to show that $r_2 \ge A_2B_3$ and $r_3 \ge A_3B_2$ and reach thereby the contradiction we were heading for. Only the first inequality will be dealt with; the argument applies mutatis mutandis to the other. Since the angle $A_2OA_3$ is not obtuse, the orthogonal projection $A'_2$ of $A_2$ on the line $A_1A_3$ falls on the closed ray $OA_3$ emanating from $O$. If $A'_2$ fell on the closed segment $B_3O$, then the image of the line $A_2A'_2$ under a slight rotation about the midpoint of the segment $A_2A'_2$ would separate the disc $\Delta_3$ and the closure $[A_0A_1A_2]$ of the triangle $A_0A_1A_2$, in contradiction with the second remark in the opening paragraph. Hence $A'_2$ lies on the open ray $B_3A_3$ emanating from $B_3$, so $\text{dist}(A_2, [A_0A_1A_3] \setminus \Delta_3) \ge A_2B_3$ by the lemma. Finally, recall that $\Delta_2$ covers points in $[A_0A_1A_3] \setminus \Delta_3$, to conclude that $r_2 \ge A_2B_3$.

Proof of the lemma. Since the quadrangle $ABCD$ is convex, the whole configuration of points lies on one side of the line $AB$, say $\mathcal{H}$. Let $F$ be the point where the ray $AD$ emanating from $A$ crosses the boundary of $\Delta$, let $\alpha$ denote the arc $EF$ of the boundary of $\Delta$ situated in $\mathcal{H}$, and let $r$ be the ray emanating from $E$ along the line $AC$, not containing $A$.
Notice that, if $X$ is a point in $[ACD] \setminus \Delta$, then the closed segment $BX$ meets either $\alpha$ or $r$ (this fails to hold if the quadrangle $ABCD$ is not convex at $D$), so it is sufficient to consider only points $X$ in $\alpha \cup r$.
Now, as a point $X$ traces $\alpha$ from $E$ to $F$, the length of the segment $BX$ varies increasingly by the cosine law in the triangle $ABX$ (this fails to hold if the quadrangle $ABCD$ is not convex at $A$ or at $B$), so $BX \ge BE$.
Finally, since the orthogonal projection of $B$ on the line $AC$ is not interior to $r$, the length of the segment $BX$ varies again increasingly, as $X$ runs along $r$ away from $E$, so $BX \ge BE$ again. This ends the proof of the lemma and completes the solution.