We can rewrite an arbitrary vector of $W$ as
\begin{align*}
\colvector{a + b - 2c\\ a + b - 2c + d\\ -2a + 2b + 4c - d\\ b + d}
& = \colvector{a\\a\\-2a\\0} +
\colvector{b\\b\\2b\\b} +
\colvector{-2c\\-2c\\4c\\0} +
\colvector{0\\d\\-d\\d}\\
&= a\colvector{1\\1\\-2\\0} +
b\colvector{1\\1\\2\\1} +
c\colvector{-2\\-2\\4\\0} +
d\colvector{0\\1\\-1\\1}
\end{align*}
Thus, we can write $W$ as
\begin{align*}
W &= \spn{\set{
\colvector{1\\1\\-2\\0},
\colvector{1\\1\\2\\1},
\colvector{-2\\-2\\4\\0},
\colvector{0\\1\\-1\\1}
}}
\end{align*}
These four vectors span $W$, but we also need to determine if they are linearly independent (turns out they are not). With an application of Theorem BS we can see that the arrive at a basis employing three of these vectors,
\begin{align*}
\begin{bmatrix}
1 & 1 & -2 & 0\\
1 & 1 & -2 & 1\\
-2 & 2 & 4 & -1\\
0 & 1 & 0 & 1
\end{bmatrix}
&\rref
\begin{bmatrix}
\leading{1} & 0 & -2 & 0\\
0 & \leading{1} & 0 & 0\\
0 & 0 & 0 & \leading{1}\\
0 & 0 & 0 &0
\end{bmatrix}
\end{align*}
Thus, we have the following basis of $W$,
\begin{align*}
B &= \set{
\colvector{1\\1\\-2\\0},
\colvector{1\\1\\2\\1},
\colvector{0\\1\\-1\\1}
}
\end{align*}