Here is my attempt:
Let $(X,\rho)$ be a metric space. By definition, a subset of a metric space is said to be closed in the metric topology if its compliment is open in $(X, \rho)$. Suppose $D_r(a)=\{x\in X : \rho(a,x)\leq r\}$ is the set of closed disks with center $a$ and radius $r$. Let $y\in X\setminus D_r(a)$. Then $\epsilon=\rho(x,y)-r>0$. Next, if $x\in D_r(y,\epsilon)$ then by the triangle inequality $\rho(x,z)\geq \rho(x,y)-\rho(y,z)>\rho(x,y)-\epsilon=r$. Which shows $(D_r(a))^\complement$ is open.
I'm usually wrong and haven't been doing amazing in this class so if someone could help me out with this I would really appreciate it. We are using the text "Elementary Topology Problem Textbook" by O.Ya. Viro, O. A. Ivanov, N. Yu. Netsvetaev, and V. M. Kharlamov and are in chapter 1 section 4 titled metric spaces.
Best Answer
First mistake: $y \in X \setminus D(a,r)$ implies $\rho (y,a )> r$. Your definition of $\epsilon$ does not make sense because you did not say what $x$ is.
Second mistake: In the last part you did not say what $z$ is.
Correct proof: $\rho (y,a )> r$. Let $\epsilon =\rho (y,a )- r$. Let us show that $B(y,\epsilon) \subset X \setminus D(a,r)$. Pick $z \in B(y,\epsilon)$. Then $\rho (z,y) <\epsilon= \rho (y,a )- r$. Hence, $\rho(z,a) \geq \rho (y,a)-\rho (y,z) >\rho (y,a)-\epsilon=r$. This finishes the proof.