The dimension of the subspace $S$ of $\Bbb R^4$ is the column rank of the matrix
$$X=\begin{bmatrix}\color{blue}1&\color{red}2&\color{green}2&\color{brown}3\\
\color{blue}2&\color{red}5&\color{green}4&\color{brown}8\\
\color{blue}{-1}&\color{red}{-3}&\color{green}{-2}&\color{brown}{-5}\\
\color{blue}0&\color{red}{2}&\color{green}0&\color{brown}4
\end{bmatrix}$$
since you have written the spanning vectors as the column vectors of the matrix and it is the column rank that gives the number of linearly independent column vectors. The column rank is found by using column transformations to obtain the column echelon form of $X$.
The dimension of the subspace $S$ of $\Bbb R^4$ is also the row rank of the matrix
$$X'=\begin{bmatrix}\color{blue}1&\color{blue}2&\color{blue}{-1}&\color{blue}0\\
\color{red}2&\color{red}5&\color{red}{-3}&\color{red}2\\
\color{green}2&\color{green}4&\color{green}{-2}&\color{green}0\\
\color{brown}3&\color{brown}8&\color{brown}{-5}&\color{brown}4
\end{bmatrix}$$
where the spanning vectors have been written as row vectors of the matrix and the row rank gives the number of linearly dependent row vectors. The row rank is found by using row transformations to obtain the row echelon form of $X'$.
You have made use of the useful property that the row rank and column rank of a matrix are identical. This means you can alternatively obtain the row echelon form of $X$, as you have done, or the column echelon form of $X'$.
Edit. As mentioned in a useful comment, the basis of the column space of $X$ consists of the column vectors of $X$ corresponding to the pivot columns in $REF(X)$. In your case, it is the first two columns, giving the basis as $\{(1,2,-1,0),(2,5,-3,2)\}$. To see why, note that if you remove the last two columns of $X$, it will still have rank $2$, meaning that the first $2$ columns are independent. Adding the third/fourth column will not change the rank.
Your work is correct but incomplete. You should show that those two vectors, $\langle 1, 0, 1, 1 \rangle$ and $\langle 0, 1, 1, -1 \rangle$ are linearly independent, as you have only shown that they span $W$. This is not too hard, as they have zeroes in different positions.
Indeed, let $a\langle 1, 0, 1, 1 \rangle + b\langle 0, 1, 1, -1 \rangle = 0$. Then
$$0 = \langle a, 0, a, a \rangle + \langle 0, b, b, -b \rangle = \langle a, b, a + b, a - b \rangle,$$
so $a = b = 0$ and they are linearly independent, and so $\dim W = 2$. Also, I'd like to point out that saying the basis is bad usage. There will be many many bases for $W$. For example, scale up both vectors by $2$. It would be more proper to say a basis.
Best Answer
I'm assuming you want to find a basis for the subspace $$S=\{(x_1,x_2,x_3,x_4) \in \mathbb{R}^4 \mid x_1+x_2+2x_3+x_4=0\;\mbox{and}\;x_1+2x_2-x_3=0\}.$$ The standard way to do this is to notice that $S$ is the kernel of the matrix $$\begin{bmatrix}1 & 1 & 2 & 1 \\ 1 & 2 & -1 & 0\end{bmatrix}.$$ Row reduce to get $$\begin{bmatrix}1 & 0 & 5 & 2 \\ 0 & 1 & -3 & -1\end{bmatrix}.$$ This tells you that a basis for $S$ is $\{(-5,3,1,0),(-2,1,0,1)\}$.