We show that for any countable subset $D$ of $\mathbb{R}^2$, there exists an open set $U$ in $(\mathbb{R}^2,\rho)$ such that $D\cap U=\emptyset$. This is enough since it will prove the non-existence of any countable dense subset of $\mathbb{R}^2$. Let $D$ be a countable set with an enumeration $D=\{x_1,x_2,\ldots,x_n,\ldots\}$. Let $R_j$ be the ray passing through $x_j$. Let $\theta_j$ be the slope of the ray $R_j$. Hence $\Theta:=\{\theta_j:j\in\mathbb{N}\}$ is an at most countable subset of $\mathbb{R}$. Hence $\mathbb{R}\setminus\Theta\neq\emptyset$. Take some $\theta\in\mathbb{R}\setminus\Theta$. Consider the ray $R_{\theta}$ corresponding to that angle $\theta$. Therefore $(R_{\theta}\setminus\{0\})\cap D=\emptyset$. We are left to show that $R_{\theta}\setminus\{0\}$ is an open set in $(\mathbb{R}^2,\rho)$. Let $r\in R_{\theta}\setminus\{0\}$. Let $t=d(r,0)>0$. Consider the open ball $B_{\rho}(r,t/2)$. Let $s$ be any point outside the ray $R_{\theta}$. Then $\rho(r,s)=d(r,0)+d(0,s)\geq d(r,0)>t/2$. Hence $s\notin B_{\rho}(r,t/2)$. Hence only points on $R_{\theta}$ belong to $B_{\rho}(r,t/2)$. Clearly $B_{\rho}(r,t/2)\subset R_{\theta}$. Hence for every point in $R_{\theta}\setminus\{0\}$, there exists an open ball in $(\mathbb{R}^2,\rho)$ completely contained in $R_{\theta}\setminus\{0\}$. Hence $R_{\theta}\setminus\{0\}$ is open in $(\mathbb{R}^2,\rho)$. Therefore $(\mathbb{R}^2,\rho)$ is not separable.
Can someone tell me if this proof is correct or there is anything wrong?
It’s basically correct, though there is one small oversight: if some $x_j$ is the origin, there isn’t a ray through $x_j$. In that case you can work with $D\setminus\{\langle 0,0\rangle\}$ instead, since you’re constructing an open set that doesn’t contain the origin anyway.
It could, however, be simplified just a bit: just observe that $\{\langle r,\theta\rangle:0<r<2\}$ (where I’m using polar coordinates) is the open ball of radius $1$ centred at $\langle 1,\theta\rangle$ and is disjoint from $D$.