variation of Vitali in $\mathbb{R}^2$ Let $B(x, r)$, be a two dimensional open disc with center $x \in \mathbb{R}^2$ and radius $r > 0$. Consider a set $E \subset \mathbb{R}^2$ such that $E \subset \bigcup_{i = 1}^n B(x_i, r_i)$ and $\sum_{i = 1}^n r_i \leq 1$ Show that then there exists a collection of non-overlapping open discs $B(y_i, R_i)$ such that $E \subset \bigcup_{i = 1}^n B(y_i, R_i)$ and $\sum_{i = 1}^n R_i \leq 1$
Clearly if the balls $B(x_i, r_i)$ are non overlapping then we are done. If not then at least $2$ balls must overlap.
First attempt
Let $B(x_1,r_1)$ be the ball whose radius is the smallest such that $B(x_1,r_1)$ overlaps with at least one other ball. Pick $B(x_j,r_j)$ whose radius is maximal among all the balls overlapping with $B(x_1,r_1)$, and note that the sums of the radius of the balls in $B(x_{j},r_{j} + r_1) \bigcup_{i = 2, i \neq j}^n B(x_i,r_i)$ is the same as the sum of the radius of the balls in $\bigcup_{i = 1}^n B(x_i, r_i)$. The problem here is that $B(x_1,r_1)$ is not necessarily contained in $B(x_j,r_j + r_1)$.
Second attempt Let $B(x_1,r_1)$ be the ball whose radius is the smallest such that $B(x_1,r_1)$ overlaps with at least one other ball. Pick $B(x_j,r_j)$ whose radius is maximal among all the balls overlapping with $B(x_1,r_1)$, and note that $B(x_1, r_1)$ is contained in $B(x_j, r_j + 2r_1)$. The problem here is that the sums of the radius of the balls in $B(x_{j},r_{j} + 2r_1) \bigcup_{i = 2, i \neq j}^n B(x_i,r_i)$ is Not the same as the sum of the radius of the balls in $\bigcup_{i = 1}^n B(x_i, r_i)$ and therefore not necessarily less then or equal to $1$.
Since I want to cover $E$ and have radius that sum up to be less then or equal to $1$, I know I want to maximize surface area while keeping the sum of the radi as small as possible so it dose feel logical to get rid of smaller circles while expanding upon larger once. Any help would be appreciated.
Edit
So after thinking about it a little longer I realized it might be impossible to create the disjoint collection of balls $B(y_i,R_i)$ such that each $y_i = x_i$ and I might have to move the discs. For example if $E$ was the union of $2$ balls of each radius $1/2$, one centered at the origin and one centered at $(7/8,0)$ then there is no way to just expand one ball without moving it so that its radius remains $1$ but it covers the other ball.
So now I am thinking start with letting $B(x_1,r_1)$ be the ball whose radius is the smallest such that $B(x_1,r_1)$ overlaps with at least one other ball. Pick $B(x_j,r_j)$ whose radius is maximal among all the balls overlapping with $B(x_1,r_1)$, Consider the ball $B( ?,r_1 + r_j)$ so I need to find a suitable value for $?$. I am temped to average the two centers $x_1, x_j$ but I know it should be closer to $x_j$, but then that makes this really complicated.
 A: I think I figured it out.
We may assume that non of the balls are covered by the other balls since if a ball was covered by other balls then we may get rid of it. Clearly if the balls $B(x_i, r_i)$ are non overlapping then we are done.
If not then at least $2$ balls must overlap. Let $B(x_1,r_1)$ and $B(x_2,r_2)$ be balls that overlap. With out loss of generality let $r_1 \geq r_2$. Draw a line segment from $x_1$ to $x_2$. Note that since  $B(x_1,r_1)$ is overlapping with $B(x_2,r_2)$, that the line segment from $x_1$ to $x_2$ can not have length more then $r_1 + r_2$. If this line segment has length at least $r_2$, then let $x'$ be the point on this line segment such that $x'$ is at a distance of $r_2$ away from $x_1$. If the distance of the line segment is shorter then let $x'$ be the midpoint of the line segment. Consider the ball $B(x',r_1 + r_2)$. Since any point in $B(x_1,r_1)$ has at most distance $r_1$ from $x_1$, and $x_1$ has distance at most $r_2$ from $x'$ we see that any point in $B(x_1,r_1)$ has at most distance $r_1 + r_2$ away from $x'$, and hence $B(x_1,r_1)$ is covered by $B(x',r_1 + r_2)$. Since the line segment is at most $r_1 + r_2$, we see that if $x'$ is at a distance of $r_2$ from $x_1$ then $x'$ must be a distance less then or equal to $r_1$ form $x_2$. If the line segment is less then $r_2$ then clearly the mid point of the line segment is less then $r_2$ away from $x_2$, and since $r_1 \geq r_2$ we get that in both cases $x'$ has distance at most $r_1$ away from $x_2$. Therefore $B(x_2, r_2)$ is also covered by $B(x',r_1 + r_2)$. Let $B'$ contain the ball  $B(x',r_1 + r_2)$ and $\cup_{i = 3}^N B(x_i,r_i)$. Note that $E$ is covered by the balls in $B'$ and that the sum of their radius remains the same as the sum of the previous collection of balls
Again ether all the balls are disjoint and we are done, or at least two balls overlap and we repeat out previous argument. Eventual we get our disjoint collection without changing the sum of the radius.
