You are correct that there does not exist a maximising metric, and the sequence you identified does saturate the upper bound. 

In order to see this, consider a slightly different problem. On $\mathbb S^2$, for $g_0$ the canonical round metric and for any Radon measure $m$ define the first non-trivial eigenvalue associated to this measure as
$$\lambda_1(m) := \inf\left\{\frac{\int_{\mathbb S^2} |\nabla_{g_0} f|^2 \, \mathrm d v_{g_0}}{\int_{\mathbb S^2} f^2 \, \mathrm d m} : f \in \mathrm C^\infty(\mathbb S^2), \int_{\mathbb S^2} f \, \mathrm d m = 0 \right\}.$$
 You can refer to Sections 3 and 4 in this paper https://link.springer.com/article/10.1007/s00039-021-00573-5 for a review on elementary properties of these eigenvalues. In particular, through a conformal change of variables you can bring the Laplace eigenvalue problem for any metric on the sphere to this form. 

Let $\phi : (\mathbb D,g) \to \mathbb (S^2,g_0)$ be a conformal map from the disk into the sphere so that $g = \phi^* g_0$, and consider $\mathrm d m = \phi_* \mathrm d v_g$ the measure on $\mathbb S^2$ obtained by  pushing forward the $g$-volume measure on the disk. Then, we have from monotonicity of the Dirichlet energy under set inclusion, as well as its conformal invariance that
\begin{align*}
\mu_1(g) &= \inf \left \{\frac{\int_{\mathbb D} |\nabla_g f|^2 \, \mathrm d v_g}{\int_\mathbb D f^2 \, \mathrm d v_g } : f \in \mathrm C^\infty(\mathbb D), \int_{\mathbb D} f \, \mathrm d v_g = 0 \right \} \\
&= \inf \left \{\frac{\int_{\phi(\mathbb D)} |\nabla_{g_0} f|^2 \, \mathrm d v_{g_0}}{\int_{\phi(\mathbb D)} f^2 \, \mathrm d m } : f \in \mathrm C^\infty(\phi(\mathbb D)), \int_{\mathbb \phi(D)} f \, \mathrm d m = 0 \right \} \\
&\le \inf \left \{\frac{\int_{\mathbb S^2} |\nabla_{g_0} f|^2 \, \mathrm d v_{g_0}}{\int_{\mathbb S^2} f^2 \, \mathrm d m } : f \in \mathrm C^\infty(\mathbb S^2), \int_{\mathbb S^2} f \, \mathrm d m = 0 \right \} \\
&= \lambda_1(m).
\end{align*}
Now let us consider the problem of maximising the functional $m \mapsto \lambda_1(m) m(\mathbb S^2) =: \bar \lambda_1(m)$, keeping in mind that our previous computation gives us that $\mu_1(g) \operatorname{Area}(g) \le \bar \lambda_1(\phi_* \mathrm d v_g)$. Now, we know that for any non-atomic Radon measure $m$, $\bar \lambda_1(m) \le 8\pi$, and that the volume measure of the round metric attains that bound. Furthermore,  by a theorem of Karpukhin and Stern (Theorem 1.4 in https://arxiv.org/pdf/2004.04086.pdf) we know that any maximal measure is does not vanish on an open set, as this would contradict unique continuation. In particular, in conjunction with your example of a sequence of metrics on the disk saturating the $8\pi$ bound, no maximal metric can exist on the disk.

In fact, that argument can be extended to show that there are never maximal metrics for any riemannian manifold with non-empty boundary, because it would imply again the existence of a maximal measure which vanishes on an open set.