Proof. $(1)$: Since $\iota: H \to E$ is continuous, $\iota^{*}: E^{*} \to H$ is continuous with respect to the strong and weak* topologies. As $H \subset E$ is dense, $\iota^{*}: E^{*} \to H^{*}$ is injective.

Let $h \in H$ such that $h \perp \iota^{*}(L)$, then by the weak*-density of $L \subset E^{*}$, $\angles{h, x^*}_{E} = 0$ for all $x^{*} \in E$. Therefore $h = 0$, and $\iota^{*}(L)$ is dense in $H$.

$(2)$: By $(1)$, the image $\iota^{*}(E^{*}) \subset H$ is dense. Hence if $x \in H$, then

On the other hand, if $\sup_{x^* \in E^*, \norm{\iota^*x^*}_H = 1}\angles{x, x^*}_{E} < \infty$ then the map $\iota^{*}(E^{*}) \to \real$ with $\iota^{*}x^{*} \mapsto \angles{x, x^*}_{E}$ is well-defined as $\iota^{*}$ is injective, and continuous by assumption. By density of $\iota^{*}(E^{*}) \subset H$, there exists $h \in H$ such that $\anglesn{h, \iota^*x^*}_{H} = \angles{h, x^*}_{E}= \angles{x, x^*}_{E}$ for all $x \in E^{*}$. In which case, $x = h \in H$.

$(3)$: Since $E^{*}$ is separable with respect to the weak* topology, there exists a linearly independent set $\seq{x_n^*}\subset L$ whose span is dense in $E$. By $(1)$, $\seq{\iota^*x_n^*}\subset H$ has dense span in $H$. Using the Gram-Schmidt process, there exists an orthonormal basis $\seq{h_n}\subset \text{span}\bracs{\iota^*x_n^*: n \in \nat}$ for $H$. Since $\iota^{*}$ is injective and $L$ is a subspace, $\bracsn{(\iota^*)^{-1}(h_n)}_{1}^{\infty} \subset L$ is the desired sequence.$\square$