Be it homework or not – this is a very fine problem indeed!
It is enough to consider a plane $E$ with points $z=(x,y)$. Let $|z|:=\sqrt{x^2+y^2}$ denote the euclidean norm.
(i) Claim: For each line $g\subset E$ there is a constant $\lambda_g$ such that for all $z$, $w\in g$ one has $d(z,w)=\lambda_g\ |z-w|\ .$ Proof. Let $g:\quad t\mapsto z(t)=a + t\ u\ ,\qquad |u|=1$ be a parametrization of $g$ with euclidean arc length as parameter and put $d^*(s,t):=d\bigl(z(s),z(t)\bigr)\ .$ Then by the basic property of $d(\cdot,\cdot)$ we have for arbitrary $s$, $t\in{\mathbb R}$: $d^*(s,t)=d^*\bigl(s,s+(t-s)\bigr)=|t-s|d^*(s,s+1)$ and similarly $d^*(s,t)=d^*(t,s)=d^*\bigl(t,t+(s-t)\bigr)=|t-s|d^*(t,t+1)\ .$ For $t:=0$ the last two equations imply $|s|d^*(s,s+1)=d^*(s,0)=|s|d^*(0,1)$. It follows that $d^*(s,s+1)=d^*(0,1)=:\lambda_g$ for arbitrary $s\in{\mathbb R}$, whence $d\bigl(z(s),z(t)\bigr)=d^*(s,t)=\lambda_g |t-s|=\lambda_g\bigl|z(t)-z(s)\bigr|\ ,$ as claimed.
(ii) Claim: The scaling factor $\lambda_g$ depends only on the direction vector $u=(\cos\phi,\sin\phi)$ of $g$. In fact there is a continuous function $\lambda:\ S^1\to{\mathbb R}_{>0}$ such that $\lambda_g= \lambda(\phi)$.
Proof. It is enough to consider the following situation: Let $g_1$ be the line $y=1$ and $g_2$ be the line $y=-1$, and let $p_i:=\lambda_{g_i}$. We have to show that $p_1=p_2$. Consider for small $\phi >0$ the line $ g_\phi:\quad t \mapsto t \ (\cos\phi,\sin\phi)\ .$ The line $g_\phi$ has a certain scaling factor $\lambda(\phi)$ and intersects $g_1$ at the point $b:=(\cot\phi, 1)$. Put $(0,0)=:o$, $(0,1)=:a$. Then we have by the triangle inequality for $d$: $d(a,b)-d(o,a)\leq d(o,b)\leq d(o,a)+d(a,b)$ or $p_1\cot\phi - d(o,a)\leq \lambda(\phi)\ {1\over\sin\phi}\leq d(o,a)+p_1\cot\phi\ .$ Multiplying by $\sin\phi>0$ we obtain $p_1 \cos\phi -d(o,a)\sin\phi \leq\lambda(\phi)\leq p_1\cos\phi+d(o,a)\sin\phi\ ,$ from which we conclude $\lim_{\phi\to 0+}\lambda(\phi)=p_1$. By symmetry (arguing on the left side) we also have $\lim_{\phi\to 0-}\lambda(\phi)=p_1$, so that in fact $\lim_{\phi\to0}\lambda(\phi)=p_1$. Using symmetry again it follows that $p_2=p_1$. Since the lines $g_1$, $g_2$ had nothing special about them it follows that any two parallel lines have the same scaling factor $\ \bigl(=:\lambda(\phi)\bigr)$; in particular the scaling factor of $g_0\colon\ y=0$ is also $\ =p_1$, which proves the continuity of the function $\lambda(\cdot)$. In particular, this function is bounded away from $0$ and $\infty$.
(iii) It follows from (i) and (ii) that the metric $d(\cdot,\cdot)$ is translation invariant. We now define the associated norm by $\|z\|\ :=\ d(o,z)\ .$ The identity $\|\alpha\ z\|=|\alpha|\ \|z\|$ is obvious, and $\|z+w\|=d(0,z+w)\leq d(o,z)+d(z,z+w)=d(o,z)+d(o,w)=\|z\|+\|w\|$ proves the triangle inequality for the norm.