We give the proof for the case when $A$ is lower triangular, and leave the case when $A$ is upper triangular for you. Consider the process for computing the inverse of a matrix that is outlined in the proof of Theorem CINM. We augment $A$ with the size $n$ identity matrix, $I_n$, and row-reduce the $n\times 2n$ matrix to reduced row-echelon form via the algorithm in Theorem REMEF. The proof involves tracking the peculiarities of this process in the case of a lower triangular matrix. Let $M=\augmented{A}{I_n}$.
First, none of the diagonal elements of $A$ are zero. By repeated expansion about the first row, the determinant of a lower triangular matrix can be seen to be the product of the diagonal entries (Theorem DER). If just one of these diagonal elements was zero, then the determinant of $A$ is zero and $A$ is singular by Theorem SMZD. Slightly violating the exact algorithm for row reduction we can form a matrix, $M^\prime$, that is row-equivalent to $M$, by multiplying row $i$ by the nonzero scalar $\matrixentry{A}{ii}^{-1}$, for $1\leq i\leq n$. This sets $\matrixentry{M^\prime}{ii}=1$ and $\matrixentry{M^\prime}{i,n+1}=\matrixentry{A}{ii}^{-1}$, and leaves every zero entry of $M$ unchanged.
Let $M_j$ denote the matrix obtained form $M^\prime$ after converting column $j$ to a pivot column. We can convert column $j$ of $M_{j-1}$ into a pivot column with a set of $n-j-1$ row operations of the form $\rowopadd{\alpha}{j}{k}$ with $j+1\leq k\leq n$. The key observation here is that we add multiples of row $j$ only to higher-numbered rows. This means that none of the entries in rows $1$ through $j-1$ is changed, and since row $j$ has zeros in columns $j+1$ through $n$, none of the entries in rows $j+1$ through $n$ is changed in columns $j+1$ through $n$. The first $n$ columns of $M^\prime$ form a lower triangular matrix with 1's on the diagonal. In its conversion to the identity matrix through this sequence of row operations, it remains lower triangular with 1's on the diagonal.
What happens in columns $n+1$ through $2n$ of $M^\prime$? These columns began in $M$ as the identity matrix, and in $M^\prime$ each diagonal entry was scaled to a reciprocal of the corresponding diagonal entry of $A$. Notice that trivially, these final $n$ columns of $M^\prime$ form a lower triangular matrix. Just as we argued for the first $n$ columns, the row operations that convert $M_{j-1}$ into $M_j$ will preserve the lower triangular form in the final $n$ columns and preserve the exact values of the diagonal entries. By Theorem CINM, the final $n$ columns of $M_n$ is the inverse of $A$, and this matrix has the necessary properties advertised in the conclusion of this theorem.