Hint: use the maximum principle for harmonic functions.
For the details, it depends a little on what tools you have available. Let me assume that you know the following theorem about removable singularities for subharmonic functions:
Theorem Assume that $\Omega$ is a domain in $\mathbb{C}$ and that $E \subset \Omega$ is polar. If $u$ is an upper bounded subharmonic function on $\Omega \setminus E$, then $u$ extends (uniquely) to a subharmonic function on $\Omega$.
From the theorem, it follows that $u$ is (can be extended to be) subharmonic on $\mathbb{D}$, and so can $-u$. Hence $u$ is in fact harmonic on $\mathbb{D}$, so by the maximum principle $u = 0$ on $\mathbb{D}$.
Just in case, here is a proof of the removable singularities theorem:
Let $\phi$ be a negative subharmonic function such that $\phi = -\infty$ on $E$. (In your particular case, you can take $\phi(z) = \log|z|$.) Let $u_\epsilon = \max \{ u, \epsilon \phi \}$. Then $u_\epsilon$ is subharmonic for every $\epsilon$, since $u_\epsilon = \phi$ near $E$. Furthermore $u_\epsilon$ increases to $u$ on $\Omega \setminus E$ as $\epsilon \to 0^+$, and the function $$v(z) = \limsup_{\zeta \to z} \left(\sup_{\epsilon > 0} u_\epsilon(z)\right)$$ is subharmonic on $\Omega$. By continuity of $u$, $v(z) = u(z)$ on $\Omega \setminus E$, and we have our subharmonic extension of $u$. (In fact, you can check that $v(z) = \limsup_{\zeta \to z} u(\zeta)$ for $z \in E$.)
