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.

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 "Continuity of eigenvalues and shape optimisation for Laplace and Steklov problems" by A. Girouard, M. Karpukhin and J. Lagacé (Geom. Funct. Anal. 31, 513–561 (2021)) 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 "Min-max harmonic maps and a new characterization of conformal eigenvalues") 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.