Ziyuang's answer handles the cases, where $N^2=0$, but it can be generalized as follows. A triangular $n\times n$ matrix $T$ with 1s on the diagonal can be written in the form $T=I+N$. Here $N$ is the strictly triangular part (with zeros on the diagonal), and it always satisfies the relation $N^{n}=0$. Therefore we can use the polynomial factorization $1-x^n=(1-x)(1+x+x^2+\cdots +x^{n-1})$ with $x=-N$ to get the matrix relation
$$
(I+N)(I-N+N^2-N^3+\cdot+(-1)^{n-1}N^{n-1})=I + (-1)^{n-1}N^n=I
$$
telling us that $(I+N)^{-1}=I+\sum_{k=1}^{n-1}(-1)^kN^k$.
Yet another way of looking at this is to notice that it also is an instance of a geometric series $1+q+q^2+q^3+\cdots =1/(1-q)$ with $q=-N$. The series converges for the unusual reason that powers of $q$ are all zero from some point on. The same formula can be used to good effect elsewhere in algebra, too. For example, in a residue class ring like $\mathbf{Z}/2^n\mathbf{Z}$ all the even numbers are nilpotent, so computing the modular inverse of an odd number can be done with this formula.
I don't know whether you are dealing with $2\times 2$ matrices or general $n \times $n$ matrices. The result is true in either case.
It may not be clear to you what these spaces are. Define addition of matrices by adding corresponding entries. So for example
$$\begin{bmatrix}
1 & 2\\
0 & 3
\end{bmatrix}
+
\begin{bmatrix}
5 & 3\\
0 & 1
\end{bmatrix}
=
\begin{bmatrix}
6 & 5\\
0 & 4
\end{bmatrix}
$$
If $c$ is a constant (a scalar, a number) then you multiply a matrix by $c$ by multiplying each entry by $c$. So for example
$$3\begin{bmatrix}
1 & 2\\
0 & 3
\end{bmatrix}
=
\begin{bmatrix}
3 & 6\\
0 & 9
\end{bmatrix}
$$
A vector space of matrices is a collection $V$ of matrices (of the same size) such that if $A$ and $B$ are matrices in the collection, then so is the sum $A+B$, and also if $c$ is any scalar, then $cA$ is in the collection.
So typically a vector space of matrices will have many matrices in it. The only vector space of matrices that consists of a single matrix is the space whose only element is the all $0$'s matrix.
In particular, the identity matrix by itself ($1$'s down the main diagonal, $0$'s elsewhere) is not a subspace of the collection of $2\times 2$ matrices, for if the identity matrix $I$ is in the subspace, then $cI$ has to be in the subspace for all numbers $c$. The collection of all matrices which are $0$ off diagonal, and have all diagonal terms equal is a subspace of the space of all matrices. Maybe that will take care of part of your objection.
Let $V$ be any vector space, and take a collection $U$ of some of the elements of $V$. Then $U$ is called a subspace of $V$ if $U$ by itself is a vector space, meaning that the sum of any two elements of $U$ is in $U$, and any constant times an element of $U$ is in $U$.
You quoted something to the effect that a certain $D$ is a subspace of the space of upper triangular matrices. That's not true. The collection of all matrices of the shape you described, with everything off diagonal equal to $0$, is a subspace. So $D$ is supposed to be not a single matrix, it is a largish collection of matrices.
Now let's look at your particular problem. Let $V$ be the collection of all upper triangular matrices. Is this a vector space? Take any two upper triangular matrices $A$ and $B$. Is $A+B$ upper triangular? Yes. If $c$ is a constant, and $A$ is upper triangular, is $cA$ upper triangular? Yes. So $V$ is a vector space.
Let $D$ be the collection of all diagonal matrices? Is this a vector space? Yes, the sum of two diagonal matrices is diagonal, a constant times a diagonal matrix is a diagonal matrix. $D$ is a subspace of the upper triangular matrices, because any diagonal matrix is in particular upper triangular, it is a special upper triangular matrix.
Best Answer
I guess the answer is 1+2+3+...+7=28. Because every element in matrices in S can be a base in that space.