As in the original question, let $\Gamma$ be the contour described by the equation $y^2-x^2=1/4$ whose upper part is oriented right-to-left and whose lower part is oriented left-to-right. We will call the region between these two curves the region interior to $\Gamma$.
Let $k > 0$ and let $C_k = \Gamma_k + \Delta_k^+ + \Delta_k^-$ be the closed contour whose components are defined by
$\begin{align} \Gamma_k &= \{x+iy \colon y^2-x^2=1/4 \text{ and } |x| \leq k\}, \\ \Delta_k^+ &= \left\{k+iy \colon |y| \leq \sqrt{k^2+1/4}\right\}, \\ \Delta_k^- &= \left\{-k+iy \colon |y| \leq \sqrt{k^2+1/4}\right\}, \\ \end{align}$
which is oriented as in the following picture.

For any $k>0$ the function $f(z) = (1+z^2)^{-1/2}$ is analytic interior to $C_k$, so Cauchy's integral formula tells us that, for $z$ interior to $C_k$,
$\begin{align} \frac{1}{(1+z^2)^{1/2}} &= \frac{1}{2 \pi i} \int_{C_k} \frac{dw}{(w-z)(1 + w^2)^{1/2}} \\ &= \frac{1}{2 \pi i} \left(\int_{\Gamma_k}g(w,z)\,dw + \int_{\Delta_k^+}g(w,z)\,dw + \int_{\Delta_k^-}g(w,z)\,dw\right), \end{align} \tag{1}$
where $g(w,z) = (w-z)^{-1} (1+w^2)^{-1/2}$. Note that for any $z$ interior to $\Gamma$, $z$ will be interior to $C_k$ when $k$ is large enough.
We will show that the integral over $\Delta_k^+$ goes to $0$ as $k \to \infty$. The calculation for the integral over $\Delta_k^-$ is identical.
Indeed, we have
$\begin{align} \left|\int_{\Delta_k^+}g(w,z)\,dw\right| &\leq L(\Delta_k^+) \cdot \sup_{w \in \Delta_k^+} |g(w,z)| \\ &\leq 2 \sqrt{k^2+1/4} \cdot \frac{1}{(k-\text{Re}(z))\sqrt{1+k^2}} \\ &= O\left(\frac{1}{k}\right). \end{align}$
Since $(1)$ holds for any $z$ interior to $\Gamma$ when $k$ is large enough, we can let $k \to \infty$. Because $\lim_{k \to \infty} C_k = \Gamma$, we get
$\begin{align} \frac{1}{(1+z^2)^{1/2}} &= \frac{1}{2 \pi i} \cdot \lim_{k \to \infty} \left(\int_{\Gamma_k}g(w,z)\,dw + \int_{\Delta_k^+}g(w,z)\,dw + \int_{\Delta_k^-}g(w,z)\,dw\right) \\ &= \frac{1}{2 \pi i} \int_{\Gamma} \frac{dw}{(w-z)(1 + w^2)^{1/2}} \end{align}$
for all $z$ interior to $\Gamma$.