Is the conjugate of a dense and continuous embedding a dense and continuous one as well?

I make a note about the title because it took a hard time finding a proof.
This topic relates to an abstract Wiener space in the Malliavin calculus.

Proposition. Let $B$ be a real separable Banach space and $H$ be a real separable Hilbert space embedded densely and continuously to $B$ . Then, $B^*$ is densely and continuously embedded to $H^*$ .

Proof. Let $\iota$ be a continuous linear injection from $H$ to $B$ such that $\iota(H)$ is dense in $B$ . In order to see that $\iota^*(B^*)$ is dense in $H^*$ , it suffices to show that for $h\in H^{**}$ , $h|_{\iota^*(B^*)}=0$ implies $h=0$ , where $\iota^*$ represents the conjugate of $\iota$ .
In what follows, $\langle\cdot,\cdot\rangle_{H}$ represents the inner product of a (general) Hilbert space $H$ and the natural bilinear form between a (general) Banach space $B$ and its conjugate $B^*$ is denoted by ${}_B\langle \cdot,\cdot \rangle_{B^*}$ . For simplicity, the conjugate space $H^*$ is identified with $H$ by the Riesz representation theorem, and corresponding elements are expressed as the same symbols.
Suppose $h\in H^{**}$ and ${}_{H^*}\langle\iota^*\ell,h\rangle_{H^{**}}=0$ for any $\ell\in B^*$ . Since ${}_{H^*}\langle\iota^*\ell,h\rangle_{H^{**}}=\langle h, \iota^*\ell \rangle_{H^*}={}_{B}\langle \iota h, \ell \rangle_{B^*}$ , the assumption ${}_{H^*}\langle\iota^*\ell,h\rangle_{H^{**}}=0$ implies
\begin{align*}
& {}_{B}\langle \iota h, \ell \rangle_{B^*} = 0. \tag{1}
\end{align*}
Now we set the continuous linear functional $\tilde{\ell}_h$ defined on $\iota(H)$ as follows:
\begin{align*}
\iota(H)\ni\iota g \mapsto \langle g, h \rangle_{H^*}\in\mathbb{R}.
\end{align*}
Since $\iota(H)$ is dense in $B$ , for any $x\in B$ , there is a sequence $(g_n)_n$ in $H$ such that $\iota g_n \to x$ as $n\to \infty$ in $B$ . Thus, we get $\ell_h\in B^*$ , by defining, for $x\in B$ , ${}_B\langle x,\ell_h\rangle_{B^*}:=\lim_{n\to\infty}\tilde{\ell}_h(\iota g_n)$ . We note that the definition of $\ell_h$ is independent of the manner to choose the sequence $(g_n)_n$ . Then, putting $\ell=\ell_h$ in (1) yields $h=0$ . $\Box$