\(\newcommand{\Cov}{\mathrm{Cov}}\) \(\newcommand{\Corr}{\mathrm{Corr}}\) \(\newcommand{\Sample}{X_{1},\dots,X_{n}}\)
Let \(\Sample\) be a random sample from a normal distribution. Then the r.v.
\begin{equation*}
\frac{(n-1)s^{2}}{\sigma^{2}}=\frac{\sum\left(x_{i}-\bar{x}\right)^{2}}{\sigma^{2}}
\end{equation*}
has a chi-squared distribution with \(n-1\) degrees of freedom.
Denote \(\chi^{2}_{\alpha,\nu}\) for the chi-squared critical value.
Do note that the chi-squared distribution is not symmetric. Also note that while the random variable depends on \(\sigma^{2}\), the distribution does not.
So:
\begin{equation*}
P\left(\chi^{2}_{1-\alpha/2,n-1}<\frac{(n-1)S^{2}}{\sigma^{2}}<\chi^{2}_{\alpha/2,n-1}\right)=1-\alpha
\end{equation*}
Thus the \(100(1-\alpha)\) CI for \(\sigma^{2}\) is:
\begin{equation*}
\frac{(n-1)S^{2}}{\chi^{2}_{\alpha/2,n-1}}<\sigma^{2}<\frac{(n-1)S^{2}}{\chi^{2}_{1-\alpha/2,n-1}}
\end{equation*}
Take the square root to get the CI for \(\sigma\).