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
The smallest subspace containing two given subspaces is their sum i.e. $S = U+V$.
A symmetric matrix will be $a_{ij} = a_{ji}$ for all $i,j$. A lower triangular matrix is one where $a_{ij} = 0$ if $i < j$.
First we will take care of what contains both:
Proof: Suppose $A$ is a matrix. Then, consider the matrix $B$ given by $B_{ij} = A_{ij}$ for $i \leq j$, such that $B$ is symmetric (which takes care of $i>j$ case). Now, for the matrix $A-B$, note that $(A-B)_{ij} = A_{ij}-B_{ij} = 0$ for $i \leq j$. Hence, $A-B$ is a lower triangular matrix. So $A = B + (A-B)$, is the sum of a symmetric and lower triangular matrix.
Hence, the answer is the entire space of matrices.
The intersection of the two subspaces consists of matrices which are lower triangular and symmetric. Because the zeros on top get reflected on the bottom due to symmetry, such a matrix can only be a diagonal matrix. It is easy to see that every diagonal matrix is symmetric and lower triangular. Hence, the answer to this is the space of diagonal matrices.