Question: Consider $\mathbb{R}^{2}$ with the topology $\mathcal{T} = \{U \subseteq \mathbb{R}^{2} : \mathbb{R}^{2} \setminus U \text{ is countable }\} \cup \{\emptyset\}$. Suppose that there is a metric $\rho$ inducing the topology $\mathcal{T}$. Choose two distinct points $x$ and $y$ and suppose that $\rho(x, y) = d$. Consider the open balls of radius $d/3$ centered at $x$ and $y$. Prove that the intersection of these balls is not empty, and that this contradicts the triangle inequality property of a metric.
My attempt: We suppose that the metric $\rho$ induces the topology $\mathcal{T}$ on $\mathbb{R}^{2}$. Then, by definition the collection of all $d$-balls $B(x, d)$ for $d > 0$ and $x \in \mathbb{R}^{2}$ is a basis for the topology $\mathcal{T}$. In particular, the balls $B(x, d/3)$ and $B(y, d/3)$ are also basis elements. By definition basis elements are also elements of the topology. By the definition of the topology, this means that $\mathbb{R}^{2} \setminus B(x, d/3)$ and $\mathbb{R}^{2} \setminus B(y, d/3)$ are countable. Then, also the union $\mathbb{R}^{2} \setminus B(x, d/3) \cup \mathbb{R}^{2} \setminus B(y, d/3)$ is countable. Now, since $\mathbb{R}^{2}$ is uncountable, it follows that the complement $\mathbb{R}^{2} \setminus (\mathbb{R}^{2} \setminus B(x, d/3)) \cup (\mathbb{R}^{2} \setminus B(y, d/3)) = B(x, d/3) \cap B(y, d/3)$ is uncountable, and thus non-empty. Now we may take a $z \in B(x, d/3) \cap B(y, d/3)$. Then $\rho(x, z) < d/3$ and $\rho(y, z) < d/3$. Since we assumed that $\rho(x, y) = d$, it follows that $d(x, y) = d > d(x, z) + d(z, y) = 2d/3$, contradicting the triangle inequality of a metric.
It follows directly that the topological space $(\mathbb{R}^{2}, \mathcal{T})$ is not metrizable, as there is no metric inducing $\mathcal{T}$.
My main concern of my proof is whether the construction I made of $B(x, d/3) \cap B(y, d/3)$ is the correct way of doing it. If anybody can help me out or give suggestions I would gladly appreciate that!