[proofplan]
The proof identifies the usual Studentized least-squares coefficient as a pivotal quantity. Under the normal linear model, this pivot has the Student $t$ distribution with $n-p$ degrees of freedom, so its probability of lying between the symmetric quantiles is $1-\alpha$. Rearranging that event gives exactly the displayed random interval for $\beta_j$.
[/proofplan]
[step:Form the Studentized coefficient pivot]
Fix $j \in \{1,\dots,p\}$ and define the random variable
\begin{align*}
T_j: \Omega &\to \mathbb{R} \\
\omega &\mapsto \frac{\hat{\beta}_j(\omega)-\beta_j}{\hat{\sigma}(\omega)\sqrt{c_{jj}}}.
\end{align*}
Since $X$ has rank $p$, the matrix $X^\top X$ is invertible and positive definite, so $C=(X^\top X)^{-1}$ is positive definite and $c_{jj}>0$. Thus $\sqrt{c_{jj}}$ is well-defined. Under the normal linear model, the Studentized coefficient statistic satisfies
\begin{align*}
T_j \sim t_{n-p}.
\end{align*}
This is the standard Student $t$ theorem for a least-squares coefficient in the normal linear model, applied with null value $\beta_{j,0}=\beta_j$; citing a result not yet in the wiki: Student $t$ distribution of the Studentized ordinary least-squares coefficient.
[guided]
We first isolate the random quantity whose distribution does not depend on the unknown parameters. For the fixed index $j$, define
\begin{align*}
T_j: \Omega &\to \mathbb{R} \\
\omega &\mapsto \frac{\hat{\beta}_j(\omega)-\beta_j}{\hat{\sigma}(\omega)\sqrt{c_{jj}}}.
\end{align*}
The denominator is meaningful because $X$ has full column rank $p$. Hence $X^\top X$ is invertible and positive definite, so its inverse $C=(X^\top X)^{-1}$ is also positive definite. In particular the diagonal entry $c_{jj}=C_{jj}$ is strictly positive, and $\sqrt{c_{jj}}$ is a positive real number.
The key probabilistic input is the standard Student $t$ theorem for a least-squares coefficient in the normal linear model: after centering $\hat{\beta}_j$ at its true value $\beta_j$ and scaling by its estimated standard error $\hat{\sigma}\sqrt{c_{jj}}$, the resulting statistic has the Student $t$ distribution with $n-p$ degrees of freedom. Applying that theorem with null value $\beta_{j,0}=\beta_j$ gives
\begin{align*}
T_j \sim t_{n-p}.
\end{align*}
This is the pivotal step because the law of $T_j$ is known and contains neither $\beta$ nor $\sigma$; citing a result not yet in the wiki: Student $t$ distribution of the Studentized ordinary least-squares coefficient.
[/guided]
[/step]
[step:Put the pivot between the symmetric Student quantiles]
Let
\begin{align*}
q := t_{n-p,1-\alpha/2}.
\end{align*}
By the definition of the $(1-\alpha/2)$-quantile and the symmetry of the Student $t_{n-p}$ distribution,
\begin{align*}
\mathbb{P}_{\beta,\sigma}(-q \leq T_j \leq q)=1-\alpha.
\end{align*}
Substituting the definition of $T_j$ gives
\begin{align*}
\mathbb{P}_{\beta,\sigma}\left(
-q \leq
\frac{\hat{\beta}_j-\beta_j}{\hat{\sigma}\sqrt{c_{jj}}}
\leq q
\right)=1-\alpha.
\end{align*}
[guided]
Set
\begin{align*}
q := t_{n-p,1-\alpha/2}.
\end{align*}
The Student $t_{n-p}$ distribution is symmetric about $0$. Therefore its lower $\alpha/2$ tail and upper $\alpha/2$ tail have equal probability:
\begin{align*}
\mathbb{P}_{\beta,\sigma}(T_j < -q)=\frac{\alpha}{2},
\qquad
\mathbb{P}_{\beta,\sigma}(T_j > q)=\frac{\alpha}{2}.
\end{align*}
Since $T_j$ has a continuous Student $t_{n-p}$ distribution, the endpoint probabilities are zero. Hence the probability that $T_j$ lies between the two symmetric quantiles is
\begin{align*}
\mathbb{P}_{\beta,\sigma}(-q \leq T_j \leq q)
=
1-\frac{\alpha}{2}-\frac{\alpha}{2}
=
1-\alpha.
\end{align*}
Now substitute the definition of the pivot:
\begin{align*}
\mathbb{P}_{\beta,\sigma}\left(
-q \leq
\frac{\hat{\beta}_j-\beta_j}{\hat{\sigma}\sqrt{c_{jj}}}
\leq q
\right)=1-\alpha.
\end{align*}
This is the coverage statement before solving the inequalities for the unknown coefficient $\beta_j$.
[/guided]
[/step]
[step:Rearrange the pivot event into the coefficient interval]
Because $\hat{\sigma}\sqrt{c_{jj}}>0$ on the event where the residual variance estimator is positive, multiplying the inequalities by $\hat{\sigma}\sqrt{c_{jj}}$ preserves their order:
\begin{align*}
-q\hat{\sigma}\sqrt{c_{jj}}
\leq
\hat{\beta}_j-\beta_j
\leq
q\hat{\sigma}\sqrt{c_{jj}}.
\end{align*}
Equivalently,
\begin{align*}
\hat{\beta}_j-q\hat{\sigma}\sqrt{c_{jj}}
\leq
\beta_j
\leq
\hat{\beta}_j+q\hat{\sigma}\sqrt{c_{jj}}.
\end{align*}
Substituting $q=t_{n-p,1-\alpha/2}$ yields
\begin{align*}
\mathbb{P}_{\beta,\sigma}\left(
\hat{\beta}_j - t_{n-p,1-\alpha/2}\hat{\sigma}\sqrt{c_{jj}}
\leq \beta_j \leq
\hat{\beta}_j + t_{n-p,1-\alpha/2}\hat{\sigma}\sqrt{c_{jj}}
\right)
=1-\alpha.
\end{align*}
Thus the displayed random interval has coverage probability $1-\alpha$, so it is a $100(1-\alpha)\%$ confidence interval for $\beta_j$.
[/step]
