Where $\gamma$ is the upper half of the unitary circle $|z|=1$.
First I noticed that $z^2 + 3$ has solutions in $\pm i\sqrt{3}$, then we have that $$ \begin{split} \int_{\gamma}\frac{1}{z^2+3} &= \int_{\gamma} \frac{1}{(z-i\sqrt{3})(z+i\sqrt{3})} dz \\ &= \frac{i}{2\sqrt{3}}\int_{\gamma}\frac{1}{z+i\sqrt{3}}dz - \frac{i}{2\sqrt{3}}\int_{\gamma}\frac{1}{z-i\sqrt{3}}dz \end{split} $$ And I found that $\left|\pm i\sqrt{3}\right| > 1$, so, those points are not bounded by $\displaystyle \gamma$, therefore, its contour integrals are equal to zero, and $$ \left|\int_{\gamma}\frac{dz}{z^2+3}\right| = |0| = 0 < \frac{\pi}{3}$$
Is my procedure/proof correct?
The path $\gamma$ is not a loop, and therefore you cannot apply Cauchy's integral formula directly. But if you define $\eta$ as $\gamma$ followed by the path that goes in a straight line from $-1$ up to $1$, then $\eta$ is a loop, and $\int_\eta\frac{\mathrm dz}{z^2+3}=0$. So,$$\int_\gamma\frac{\mathrm dz}{z^2+3}=-\int_{-1}^1\frac{\mathrm dz}{z^2+3}.$$And$$0<\int_{-1}^1\frac{\mathrm dz}{z^2+3}=\left[\frac1{\sqrt3}\arctan\left(\frac x{\sqrt3}\right)\right]_{x=-1}^{x=1}=\frac\pi{3\sqrt3}<\frac\pi3.$$