Part 1): Show that the product of two upper triangular matrices is upper triangular.
Let $A, B \in M_n(\mathbb{R})$ be upper triangular matrices (with respect to the standard basis). Then we can express them as
$$\left( \begin{array}{ccc}
a_{11} & a_{12} & \cdots &a_{1n} \\
0 & a_{22} & \cdots & a_{2n} \\
\vdots & 0 & \cdots & \vdots \\
0 & \cdots & 0 & a_{nn}\end{array} \right) \text{ and }\left( \begin{array}{ccc}
b_{11} & b_{12} & \cdots &b_{1n} \\
0 & b_{22} & \cdots & b_{2n} \\
\vdots & 0 & \cdots & \vdots \\
0 & \cdots & 0 & b_{nn}\end{array} \right)$$
Hence their product is clearly
$$\left( \begin{array}{ccc}
a_{11} & a_{12} & \cdots &a_{1n} \\
0 & a_{22} & \cdots & a_{2n} \\
\vdots & 0 & \cdots & \vdots \\
0 & \cdots & 0 & a_{nn}\end{array} \right)\left( \begin{array}{ccc}
b_{11} & b_{12} & \cdots &b_{1n} \\
0 & b_{22} & \cdots & b_{2n} \\
\vdots & 0 & \cdots & \vdots \\
0 & \cdots & 0 & b_{nn}\end{array} \right) = \left( \begin{array}{ccc}
a_{11}b_{11} & a_{11}b_{12}+a_{12}b_{22} & \cdots &\sum_{i=1}^n a_{1i}b_{in} \\
0 & a_{22}b_{22} & \cdots & \sum_{i=2}^n a_{2i}b_{in} \\
\vdots & 0 & \cdots & \vdots \\
0 & \cdots & 0 & a_{nn}b_{nn}\end{array} \right)$$
I think I have given you enough information to describe the general pattern of entry $c_{jk}$ of the product matrix.
Part 2): To do this, show that triangular matrices are closed under linear combinations. In particular, show that (1) the zero matrix is technically upper triangular, (2) that any triangular matrix times a scalar is upper triangular, and (3) that the sum of any two upper triangular matrices is itself upper triangular. These should be very straightforward proofs. See here for a specific case of the dimension argument.
Part 3): If $M(\phi, B)$ is the matrix representation of $\phi$ and it is upper triangular, then this means that $\phi(e_k)$ is the first column and has all zero entries except for the first $k$. Stop and think about what this means in terms of the subspace of $E$ that contains $\phi(e_k)$. I am being intentionally vague because this is an important concept that you should reach on your own, but I can be more explicit and supply a reference need be.
Part 4): By reversible do you mean invertible? If so, just think about rearranging the basis vectors after inverting the map.
While I prefer the original proof, here is an attempt by (ab)using the Jordan normal form.
Let $A = S J S^{-1}$ be a Jordan decomposition of $A$. Denote generalized eigenvectors (columns of $S$) as $s_i$ and orthonormalize them in that order, i.e., write them as
$$s_j = \sum_{i=1}^j \xi_{ij} u_i,$$
where $u_1,\dots,u_n$ are orthonormal. Let $U = \begin{bmatrix} u_1 \dots u_n \end{bmatrix}$ be a matrix with columns $u_i$, and let $X = \begin{bmatrix} \xi_{ij} \end{bmatrix}$ be a matrix with elements $\xi_{ij}$ (we define $\xi_{ij} = 0$ for $i > j$, so $X$ is upper triangular). Then
$$S = UX,$$
so
$$A = S J S^{-1} = U X J X^{-1} U^{-1} = U T U^*,$$
where $T = X J X^{-1}$ is upper triangular, because the inverse $X^{-1}$ of an upper triangular matrix is upper triangular, and the product of upper triangular matrices $X$, $J$, and $X^{-1}$ is upper triangular.
Best Answer
Consider the zero matrix. It is upper triangular, the determinant is zero.