Showing that a metric space $(\mathbb R^2, d)$ is complete Let $d: \mathbb R^2 \times \mathbb R^2 \rightarrow \mathbb R^+_0$ be the distance function with
$$ d(x, y) =\begin{cases} \big|\|y\| − \|x\|\big| & \text{if }x = \lambda y \text{ or }y = \lambda x \text{ for some } \lambda > 0, \\ 
\|x\| + \|y\| &\text{otherwise}.\end{cases}$$
Is this space $(\mathbb R^2, d)$ complete?
My idea of showing this would be to look for a Cauchy sequence $(a_n)_{n \in \mathbb N}$ and consider the cases $\lim \inf_{ n\rightarrow \infty}\|a_n\| > 0$ and $= 0$ but I'm not making much progress.
 A: Your idea for considering those two cases is good, but you still have to dig into an analysis of the two cases.
But before I do that, it helps to have some intuition. First, the metric restricted to each ray $\mathbb R^+_0 \cdot a$ (for each $a \ne (0,0)$) is identical to the ordinary Euclidean metric on that ray, and in particular the metric restricted to that ray is complete. Geometrically one many note that if two points lie on the same ray then the shortest path connecting them is the line segment on that ray; whereas if two points do not lie on the same ray, the shortest path connecting them is the path starting from one of the points, going back along the line segment connecting that point to the origin, and then going out along the path connecting the origin to the other point. So one might think of this as the "ray metric" on the plane.
Consider first the case where $L = \liminf_{n \to \infty} \|a_n\| > 0$. The thing to notice is that there exists $N > 0$ such that for all $n \ge N$ we have $\|a_n\| > L/2$ and for all $m,n \ge N$ we have $d(a_m,a_n) < L/4$. From this it immediately follows that all the points $a_n$ with $n \ge N$ lie on the same ray $\mathbb R^+_0 \cdot a_N$. That ray is a complete metric space, and $(a_n)_{n \ge N}$ is a Cauchy sequence in this ray, so it converges. Hence, the entire sequence $(a_n)_{n \ge 1}$ converges.
Consider next the case where $L = \liminf_{n \to \infty} \|a_n\| = 0$. We'll prove that $(a_n)$ converges to the origin. First, using that $L=0$ it follows that there is a subsequence $(a_{n_i})$ such that $\lim_{i \to \infty} \|a_{n_i}\|=0$, and therefore that subsequence converges to the origin. Given $\epsilon > 0$, choose $N$ so that if $m,n \ge N$ then $d(a_m,a_n) < \epsilon/2$. Then choose a single $i > 0$ so that $n_i > N$ and $\|a_{n_i}\| < \epsilon/2$. It follows that if $m \ge N$ then $d(a_m,a_{n_i}) < \epsilon/2$. In each of the two cases of the definition of $d(\cdot,\cdot)$, you can conclude that $\|a_m\| < \epsilon$.
