Let $\beta\mathbb{N}$ be the set of nonzero multiplicative linear functionals on $\ell_{\infty} (=\ell_{\infty}(\mathbb{N},\mathbb{C}))$ with the weak$^{*}$ topology. For any $n\in\mathbb{N}$, $\rho_{n}(f)=f(n)$ defines an element $\rho_{n}$ in $\beta\mathbb{N}$, and we denote this subset of $\beta\mathbb{N}$ by $\mathbb{N}$. It is not hard to see that $\beta\mathbb{N}$ is a compact Hausdorff space and that the map $f\mapsto \hat{f}$ defines an isometric isomorphism of $\ell_{\infty}$ onto the subspace $\widehat{\ell_{\infty}}$ of $C(\beta\mathbb{N})$ (where $\hat{f}(\rho)=\rho(f)$ for each $\rho\in\beta\mathbb{N}$). What I need to show is that $\widehat{\ell_{\infty}}$ is norm closed inside $C(\beta\mathbb{N})$ $\textit{without}$ using the fact that $\mathbb{N}$ is dense in $\beta\mathbb{N}$.
Assume $||\hat{f_{n}}-g||\rightarrow0$ for some sequence $\{f_{n}\}\subset \ell^\infty$ and $g\in C(\beta\mathbb{N})$. Then $|\hat{f_{n}}(\rho)-g(\rho)|\rightarrow0$ for each $\rho\in\beta\mathbb{N}$, and hence $\{\hat{f_{n}}(\rho)\}$ is cauchy. Taking $\rho$ to be $\rho_{k}$ for each $k\geq1$, we have $\{f_{n}(k)\}$ converges as $n\rightarrow\infty$ for each $k$. Defining $h(k)=\lim f_{n}(k)$ gives an element $h\in\ell^{\infty}$ (since $\{||f_{n}||\}=\{||\hat{f_{n}}||\}$ is uniformly bounded). So we get that $g(\rho_{k})=\hat{h}(\rho_{k})$ for each $k$. This is where I am stuck.
In general, if $E, F$ are Banach spaces and $T\colon E\to F$ is isometric, then $T(E)$ is closed:
If $(y_n)$ is a Cauchy sequence in $T(E)$, then $(T^{-1}y_n)$ is Cauchy as well since $T$ is isometric. Hence there exists $x\in E$ such that $T^{-1}y_n\to x$. Using again that $T$ is isometric, we obtain $y_n\to Tx$.