Let $\mathcal{A}$ be a unital $\mathrm{C}^*$-algebra and $\varphi\in\mathcal{S}(\mathcal{A})$ a state. Define the $\phi$-variance of self-adjoint $a\in\mathcal{A}$ by: $$\sigma_\phi(a)^2=\phi((a-\phi(a)1_{\mathcal{A}})^2)=\phi(a^2)-\phi(a)^2.$$
Fix $\phi\in\mathcal{S}(\mathcal{A})$ and let $\omega_\phi\in\mathcal{S}(\mathcal{A}^{**})$ be its extension to the bidual. Denote by $\mathbf{1}_E(a)\in\mathcal{A}^{**}$ the spectral projections of $a\in\mathcal{A}$. Suppose that for $\lambda\in \sigma(a)$ and for all $\varepsilon>0$, where $B_\varepsilon=(\lambda-\varepsilon,\lambda+\varepsilon)$ and $p_\varepsilon:=\mathbf{1}_{B_\varepsilon}(a)$: $$\omega_\phi(p_\varepsilon)>0.$$ Then for each $\varepsilon>0$ define $\phi_\varepsilon\in\mathcal{S}(\mathcal{A})$: $$\phi_\varepsilon(b)=\dfrac{\omega_\phi(p_\varepsilon b p_\varepsilon)}{\omega_\phi(p_\varepsilon)}\qquad(b\in\mathcal{A}).$$ (Note that the limit $\lim_{\varepsilon\to 0}\phi_\varepsilon(b)$ does not exist in general).
I am saying something here like "$\phi_\varepsilon$ $\varepsilon$-localises $a$ near $\lambda\in\sigma(a)$."
Given these conditions, I am hoping to bound $\sigma_{\phi_\varepsilon}(a)\leq f(\lambda,\varepsilon)$. I have a proposition below, but I am concerned about how I am handling the extension of $\varphi$ to $\omega_\varphi$, where I am situating elements in a closure of $C(\sigma(a))$ in something like $L^\infty(\sigma(a),\mu_\phi)$, where $\mu_\phi$ is coming from functional calculus, and there is an unresolved tension here considering I am using the bidual..
Proposition: Given the conditions above, $$\sigma_{\phi_\varepsilon}(a)^2\leq 4\varepsilon|\lambda|.$$
'Proof' Fix $\varepsilon>0$ and let $\varphi:=\phi_\varepsilon$. If we consider the $\mathrm{C}^*$-subalgebra $\mathrm{C}^*(a)$ generated by $a\in\mathcal{A}$ we have $\mathrm{C}^*(a)\cong C(\sigma(a))$, and the restriction of any state $\phi\in\mathcal{S}(\mathcal{A})$ is given by integration against a Radon probability measure $\mu_\phi$: $$\phi(b)=\int_{\sigma(a)}\pi(b)(t)\,d\mu_\phi(t).$$ Here $\pi:\mathrm{C}^*(a)\to C(\sigma(a))$ is transformation, and under this transform $a\mapsto [t\mapsto t]$.
By assumption, $\omega_\phi(p_\varepsilon)>0$. I want to say that: $$\omega_\phi(p_\varepsilon)=\int_{\sigma(a)}\mathbf{1}_{B_\varepsilon}(t)\,d\mu_\phi(t)=\mu_{\phi}(B_\varepsilon)>0.$$
I also want to say that:
$$\omega_\phi(p_\varepsilon a p_\varepsilon)=\int_{B_\varepsilon} t\,d\mu_\phi(t),$$
which yields the bounds:
$$ \begin{aligned} (\lambda-\varepsilon)\mu_\phi(B_\varepsilon)\leq &\, \omega_\phi(p_\varepsilon a p_\varepsilon)\,\leq (\lambda+\varepsilon)\mu_\phi(B_\varepsilon) \\\implies (\lambda-\varepsilon)&\leq\varphi(a)\leq (\lambda+\varepsilon)\ \end{aligned}$$ Similar methods, with $\pi(a^2)=[t\mapsto t^2]$ then yields: $$\begin{aligned} (\lambda-\varepsilon)^2&\leq\varphi(a^2)\leq (\lambda+\varepsilon)^2 \\ \implies\sigma_{\phi_\varepsilon}^2=\sigma_\varphi(a)^2&=\varphi(a^2)-\varphi(a)^2 \\ &\leq(\lambda+\varepsilon)^2-(\lambda-\varepsilon)^2=4\varepsilon\lambda. \end{aligned}.$$
This is for $\lambda>0$. For negative we end up with the absolute value. We need $\varepsilon<|\lambda|$.
For $\lambda=0$, it is
$\sigma_\varphi(a)^2\leq \varepsilon^2$.
Question: Does this analysis check out?
Here are my comments, summarized into an answer:
Everything is fine. Indeed, one could just work in $\mathcal{A}^{∗∗}$ from the very start. Then the question becomes when $\mathcal{A}$ is a vNa and $\phi$ is a normal state, whether the argument works out. You never need to mention the double dual in that case, and everything checks out.
For technical details on the Borel functional calculus, if the vNa in question is separable, then the image of Borel functional calculus is an $L^\infty(\sigma(a))$. But this doesn't always work in the non-separable case. People usually only consider the separable case, but since we're dealing with double duals, that's usually not separable. However, Borel functional calculus still works even without that identification. Indeed, if your vNa acts on Hilbert space $H$, then for each pair of vectors $h, k \in H$, $C(\sigma(a)) \ni f \mapsto \langle h, f(a)k \rangle \in \mathbb{C}$ defines a complex measure $\mu_{h, k}$ on $\sigma(a)$. Therefore, for each bounded Borel function $f$ on $\sigma(a)$, $(h, k) \mapsto \int f d\mu_{h, k}$ defines a bounded sesquilinear form on $H$ and therefore induces an operator $f(a)$. This is how Borel functional calculus is usually defined in the first place. Noting that any normal state $\phi$ (emphasis on normal) is induced by a trace-class operator easily tells you $\phi(f(a)) = \int f d\mu_\phi$. Or, if you want, you can just choose the universal normal representation of your vNa, then any normal state is a vector state and this just becomes part of the definition of Borel functional calculus.
For the bounds on integrals, the original bounds on the integrals only work when $\lambda > 0$ and $\varepsilon < \lambda$. If $\lambda < 0$ and $\varepsilon < |\lambda|$, your bounds should become $|\varphi(a)| \geq |\lambda| − \varepsilon$ and $\varphi(a^2) \leq (|\lambda| + \varepsilon)^2$, so $\varphi(a^2) − \varphi(a)^2 \leq 4\varepsilon|\lambda|$. If $\lambda = 0$, your bounds should become $|\varphi(a)| \geq 0$ and $\varphi(a^2) \leq \varepsilon^2$, so $\varphi(a^2) − \varphi(a)^2 \leq \varepsilon^2$.