Covering each point of the plane with circles three times

Below I construct a cover of $\mathbb{R}^2$ with degree $3$. The idea is to build an almost-"degree 2 cover" of $\mathbb{R}^2$ for which a single point is covered three times (using variations on the trick from the degree 2 cover given in the question), then to balance things out by adding the degree $1$ cover of $\mathbb{R}^2 - \{O\}$.

Lemma: Let $U$ be an open disk and let $P$ be a point on its boundary. Then there is a family $\mathscr{F}$ of circles contained in $U \cup \{P\}$ such that each point of $U$ belongs to exactly two circles of $\mathscr{F}$, while $P$ belongs to exactly one circle of $\mathscr{F}$.

Proof: For $a > 0$, let $L(a)$ be the line given by the equation $x = a$, and let $C(a)$ be the circle of radius $a$ centered at $(a, 0)$, so $C(a)$ passes through the origin $O$. Assume without loss of generality that $P = O$ and that $U$ is the open disk bounded by $C(1/2)$. Now, for $0 < a < b$, let $\mathscr{G}(a, b)$ be the family of circles of radius $\frac{b-a}{2}$ with centers on the line $L(\frac{b+a}{2})$, so the circles of $\mathscr{G}(a, b)$ are contained within the closed strip bounded by $L(a)$ and $L(b)$, and each point of $L(a)$ and $L(b)$ belongs to exactly one circle of $\mathscr{G}(a, b)$, while each point in the interior of the strip belongs to exactly two circles of $\mathscr{G}(a, b)$. Define $$\mathscr{G} = \bigcup_{n=1}^\infty \mathscr{G}\left(1 + \frac{1}{n+1}, 1 + \frac{1}{n} \right) \cup \bigcup_{n=1}^\infty \mathscr{G}\left(2 + \frac{1}{n+1}, 2 + \frac{1}{n} \right) \cup \bigcup_{n=3}^\infty \mathscr{G}(n, n+1)$$ so that, if we define $V = \bigcup_{a > 1} L(a) = \{(x, y) : x > 1\}$, all circles in $\mathscr{G}$ are contained in $V$, and each point in $V$ belongs to exactly two circles in $\mathscr{G}$, except for those points in $L(2)$, each of which belongs to exactly one circle in $\mathscr{G}$. Now, under inversion with respect to the unit circle centered at the origin, $L(a)$ maps to $C(\frac{1}{2a}) - \{O\}$, $V$ maps to $U$, and each circle in $\mathscr{G}$ maps to a circle in a new family $\mathscr{G}'$ of circles contained in $U$. By the above, each point in $U$ belongs to exactly two circles in $\mathscr{G}'$, except for those points on $C(1/4) - \{O\}$, each of which belongs to exactly one circle in $\mathscr{G}'$. To conclude, we define $\mathscr{F} = \mathscr{G}' \cup \{C(1/4)\}$, which has the desired properties.

Proposition: There is a cover of $\mathbb{R}^2$ with degree $3$.

Proof: Define $C'(r)$ to be the circle of radius $r$ centered at the origin. Let $P = (2, 0)$ and let $U$ be the open disk bounded by $C'(2)$. For each $n \geq 1$, let $\mathscr{H}_n$ be the family of unit circles with centers on $C'(2n+1)$, so if we let $E_n$ be the interior of the region bounded by $C'(2n)$ and $C'(2n+2)$, then each circle in $\mathscr{H}_n$ is contained in $C'(2n) \cup E_n \cup C'(2n+2)$, and each point in $E_n$ belongs to exactly two circles in $\mathscr{H}_n$, while each point on $C'(2n)$ or $C'(2n+2)$ belongs to exactly one circle in $\mathscr{H}_n$. Letting $\mathscr{H} = \bigcup_{n=1}^\infty \mathscr{H}_n$, we see that each point in $\mathbb{R}^2 - (C'(2) \cup U)$ belongs to exactly two circles in $\mathscr{H}$, each point in $C'(2)$ belongs to exactly one circle in $\mathscr{H}$, and each point in $U$ belongs to zero circles in $\mathscr{H}$. To cover $U$, let $\mathscr{F}$ be the family of circles given by the Lemma applied to $U$ and $P$, and define $\mathscr{F}' = \mathscr{F} \cup \mathscr{H} \cup \{C'(2)\}$, so each point of $\mathbb{R}^2$ belongs to exactly two circles in $\mathscr{F}'$, except $P$ which belongs to exactly three (since it lies on $C'(2)$). But none of the circles in $\mathscr{F}'$ are centered at $P$, so we can extend the family $\mathscr{F}'$ to a covering of $\mathbb{R}^2$ with degree $3$ by adding all circles centered at $P$.