$C$-doubling $\Bbb R^2$ measure gives measure zero to a straight line? A metric space $X$ with metric $d$ is said to be doubling on $\Bbb R^2$ if there is some constant $C > 0$ such that for any $x \in X$ and $r > 0$, the Euclidean ball $B(x, r) = \{y:|x − y| < r\}$ may be contained in a union of no more than $C$ many balls with radius $r/2$. 
That is:
$$B(x, 2r) < C \cdot B(x,r).$$
Can we prove that any such measure gives measure zero to a straight line $L$?
 A: Although the question was solved by the OP, there is no solution here for others to read. Let's fix that. 
Definition. A set $E\subset \mathbb R^n$ is porous if there exists $c>0$ such that for any ball $B\subset\mathbb R^n$ the set $B\setminus E$ contains a ball of diameter $c\,\mathrm{diam}\, (B)$.
Observe that a line in $\mathbb R^2$ is porous ($c=1/2$ works).
Definition. A Radon measure $\mu$ on $\mathbb R^n$ is doubling if there exists $C$ such that $\mu(2B)\le C\,\mu (B)$ for any ball $B$. An equivalent form of the definition involves squares: two squares with a common side get about the same measure, up to a multiplicative constant.
Claim. If $\mu$ is a doubling measure and $E$ is a porous set, then $\mu(E)=0$. 
Proof. It suffices to show that $\mu(E\cap Q)=0$ for every square $Q$. To begin with, note that $\mu(Q)$ is finite. Divide $Q$ into $N^2$ equal subsquares, where $N$ is something like $100/c$; this will make sure that at least one of the subsquares lies in the complement of $E$. The doubling property implies that $\mu$ is distributed somewhat fairly between subsquares: namely, each subsquare gets at least $\epsilon \,\mu(Q)$ where $\mu$ depends only on $N$ and on the doubling constant of the measure. Removing a subsquare that lies outside of $E$ from consideration, we are left with $N^2-1$ subsquares with total measure at most $(1-\epsilon)\,\mu(Q)$. Repeating this divide-and-remove process, we obtain $\mu(E\cap Q)\le (1-\epsilon)^k\,\mu(Q)$ for every $k$. Thus $\mu(E\cap Q)=0$. $\quad \Box$
This just in! A doubling measure on $\mathbb{R}^d$ can charge a rectifiable curve. 
