If $U$ is twice-differentiable at the point $x_0$ and $U''(x_0)\neq 0$, then there exists a function $R$ such that
$$U(x) = U(x_0) + U'(x_0) (x-x_0) +\frac{1}{2} U''(x_0)(x-x_0)^2 + R(x)(x-x_0)^2$$
and $\lim_{x\rightarrow x_0} R(x) = 0$. Given that $x_0$ is taken to be a stable equilibrium point of $U$, we have that $U'(x_0)= 0$ and $U''(x_0)>0$. We therefore have that
$$\sqrt{E-U(x)} = \sqrt{E-E_0 - \frac{1}{2} U''(x_0)(x-x_0)^2 - R(x)(x-x_0)^2}$$
where in accordance with OP's notation we define $U(x_0) \equiv E_0$. For $E>E_0$, we may define $\epsilon \equiv E-E_0$ and write
$$\sqrt{E-U(x)} = \sqrt{\epsilon} \sqrt{1- \frac{1}{2\epsilon}\big[U''(x_0)+2R(x)\big] (x-x_0)^2}$$
From this it follows that we may write
$$\frac{\mathrm dx}{\sqrt{E-U(x)}}= \frac{dx/\sqrt{\epsilon}}{\sqrt{1-\frac{1}{2}\big[U''(x_0)+2R(x)\big]\left(\frac{x-x_0}{\sqrt{\epsilon}}\right)^2}}$$
$$= \frac{d\xi}{\sqrt{1-\frac{1}{2}\big[U''(x_0)+2R\left(\xi\sqrt{\epsilon}+x_0\right)\big] \xi^2}}$$
where we've defined $\xi \equiv (x-x_0)/\sqrt{\epsilon}$. The turning points between which we are integrating occur when $U(x)=E$, which can be rewritten implicitly as
$$\xi= \pm\frac{1}{\sqrt{\frac{1}{2}\big[U''(x_0)+2R\big(\xi\sqrt{\epsilon}+x_0\big)\big] }}$$
Let the solutions to this equation be $\xi_1$ and $\xi_2$, respectively. Our integral then becomes
$$T_{\epsilon} = \frac{2}{\sqrt{2}}\int_{\xi_1}^{\xi_2} \frac{\mathrm d\xi}{\sqrt{1-\frac{1}{2}\big[U''(x_0)+2R\left(\xi\sqrt{\epsilon}+x_0\right)\big] \xi^2}}$$
$\epsilon$ appears in the argument of $R$ as well as in the integral bounds, so taking the $\epsilon\rightarrow 0$ limit is a bit subtle - however, I claim that we obtain the correct answer by naively taking all three limits simultaneously (see below for a more rigorous justification). If we do so, we may note that $\lim_{\epsilon\rightarrow 0} R\big(\xi\sqrt{\epsilon}+x_0\big) = 0$ and $\lim_{\epsilon\rightarrow 0} \xi_{1/2} = \pm \xi_0$ where $\xi_0 \equiv \sqrt{2/U''(x_0)}$. Plugging all of this in, we obtain
$$\lim_{\epsilon\rightarrow 0} T_\epsilon =\frac{2}{\sqrt{2}}\int_{-\xi_0}^{\xi_0} \frac{1}{\sqrt{1-\frac{1}{2}U''(x_0) \xi^2}}$$
We conclude in the obvious way - define $\sigma \equiv \xi \sqrt{U''(x_0)/2}$ to obtain
$$\lim_{\epsilon\rightarrow 0}T_\epsilon = \frac{2}{\sqrt{U''(x_0)}}\int_{-1}^1 \frac{d\sigma}{\sqrt{1-\sigma^2}} = \frac{2\pi}{\sqrt{U''(x_0)}}$$
For those interested, I'll justify my naive limit-taking more rigorously. Let $\{\epsilon_n\}$ and $\{\alpha_n\}$ be two positive sequences tending to zero such that $0<\epsilon<\epsilon_n \implies |R\big(\xi\sqrt{\epsilon} + x_0\big)|<\alpha_n$ for all each $\xi\in[\xi_1,\xi_2]$.
Define $\gamma_n = \sqrt{\frac{2}{U''(x_0)+2\alpha_n}}$ and observe that $\xi_1 < -\gamma_n$ and $\gamma_n < \xi_2$. We may now define the sequence
$$t_n = \frac{2}{\sqrt{2}}\int_{-\gamma_n}^{\gamma_n} \frac{\mathrm d\xi}{\sqrt{1-\frac{1}{2}\big[U''(x_0)+2R\big(\xi\sqrt{\epsilon_n}+x_0\big)\big]\xi^2}}$$
and note that $\lim_{n\rightarrow \infty} t_n = \lim_{\epsilon\rightarrow 0}T_\epsilon$.
Note that the integrand is bounded below by $\left(1-\frac{1}{2}\big[U''(x_0)-2\alpha_n\big]\right)^{-1/2}$. The integral of this lower bound is
$$L_n = \frac{2}{\sqrt{2}}\int_{-\gamma_n}^{\gamma_n} \frac{\mathrm d\xi}{\sqrt{1-\frac{1}{2}\big[U''(x_0)-2\alpha_n\big]\xi^2}}$$
$$= \frac{4}{\sqrt{U''(x_0)-2\alpha_n}}\sin^{-1}\left(\sqrt{\frac{U''(x_0)-2\alpha_n}{U''(x_0)+2\alpha_n}}\right)$$
Similarly, the integrand is bounded above by $\left(1-\frac{1}{2}\big[U''(x_0)+2\alpha_n\big]\right)^{-1/2}$, and the integral by
$$U_n = \frac{2}{\sqrt{2}}\int_{-\gamma_n}^{\gamma_n} \frac{\mathrm d\xi}{\sqrt{1-\frac{1}{2}\big[U''(x_0)+2\alpha_n\big]\xi^2}} = \frac{2\pi}{\sqrt{U''(x_0)+2\alpha_n}}$$
Because $L_n < t_n < U_n$ and $\lim_{n\rightarrow \infty} L_n = \lim_{n\rightarrow \infty} U_n = \frac{2\pi}{\sqrt{U''(x_0)}}$, we have by the squeeze theorem that $\lim_{n\rightarrow \infty} t_n = \lim_{\epsilon\rightarrow 0} T_\epsilon = \frac{2\pi}{\sqrt{U''(x_0)}}$.
