Why is $BB^T$ always invertible?

The statement is not true.

For example if $A= \begin{bmatrix} 2& 3\\1 & 1 \end{bmatrix}$,then $det(BB^T)=0$ so $BB^T$ is not invertible.

For a general matrix $B$ this statement is not true, see the example above. But for the Karmarkar algorithm or interior point method you specified the $B$.

If you want to solve the following LP $min \ c^Tx \quad s.t. Ax = 0, e^Tx = n, x \geq 0$ where $x \in \mathbb{R}^n, A \in \mathbb{R}^{m \times n}$ with $n>2, m<n$, $e = (1,...,1)^T \in \mathbb{R}^n$ and with $ rank A = m, Ae = 0$. Let $\bar{x}$ be a feasible point then we define $D = diag(\bar{x})$. The matrix $B$ is defined by $$B = \begin{pmatrix} AD \\ e^T\end{pmatrix} \in \mathbb{R}^{(m+1) \times n}$$ Thus $$BB^T \in \mathbb{R}^{(m+1)\times (m+1)}$$ Then you only have to show that $$rank (B) = m+1$$ since $rank(A) = rank(AA^T)$

To show this we look at the kernel of $B$ and show the kernel is only the zero vector. Let $$0 = B^T\begin{pmatrix} z \\ z_{m+1} \end{pmatrix} = ADz + z_{m+1}e$$ Since $Ae = 0$ we have that $ADA^T z = 0$ since $rank(A) = m$ and $D$ is a positive diagonal matrix. So we have $z = 0$ and it follows that $z_{m+1}$ also has to be zero. Thus we have $$rank(B) = m+1$$ and $BB^T$ has always an inverse under the constraints above.