The null space of $A$ is the set of solutions to $A{\bf x}={\bf 0}$. To find this, you may take the augmented matrix $[A|0]$ and row reduce to an echelon form. Note that every entry in the rightmost column of this matrix will always be 0 in the row reduction steps. So, we may as well just row reduce $A$, and when finding solutions to $A{\bf x}={\bf 0}$, just keep in mind that the missing column is all 0's.
Suppose after doing this, you obtain
$$
\left[\matrix{1&0&0&0&-1 \cr 0&0&1&1&0 \cr 0&0&0&0&0 \cr 0&0&0&0&0 \cr }\right]
$$
Now, look at the columns that do not contain any of the leading row entries. These columns correspond to the free variables of the solution set to $A{\bf x}={\bf 0}$ Note that at this point, we know the dimension of the null space is 3, since there are three free variables. That the null space has dimension 3 (and thus the solution set to $A{\bf x}={\bf 0}$ has three free variables) could have also been obtained by knowing that the dimension of the column space is 2 from the rank-nullity theorem.
The "free columns" in question are 2,4, and 5. We may assign any value to their corresponding variable.
So, we set $x_2=a$, $x_4=b$, and $x_5=c$, where $a$, $b$, and $c$ are arbitrary.
Now solve for $x_1$ and $x_3$:
The second row tells us $x_3=-x_4=-b$ and the first row tells us $x_1=x_5=c$.
So, the general solution to $A{\bf x}={\bf 0}$ is
$$
{\bf x}=\left[\matrix{c\cr a\cr -b\cr b\cr c}\right]
$$
Let's pause for a second. We know:
1) The null space of $A$ consists of all vectors of the form $\bf x $ above.
2) The dimension of the null space is 3.
3) We need three independent vectors for our basis for the null space.
So what we can do is take $\bf x$ and split it up as follows:
$$\eqalign{
{\bf x}=\left[\matrix{c\cr a\cr -b\cr b\cr c}\right]
&=\left[ \matrix{0\cr a\cr 0\cr 0\cr 0}\right]+
\left[\matrix{c\cr 0\cr 0\cr 0\cr c}\right]+
\left[\matrix{0\cr 0\cr -b\cr b\cr 0}\right]\cr
&=
a\left[ \matrix{0\cr1\cr0\cr 0\cr 0}\right]+
c\left[ \matrix{1\cr 0\cr 0\cr 0\cr 1}\right]+
b\left[ \matrix{0\cr 0\cr -1\cr 1\cr 0}\right]\cr
}
$$
Each of the column vectors above are in the null space of $A$. Moreover, they are independent. Thus, they form a basis.
I'm not sure that this answers your question. I did a bit of "hand waving" here. What I glossed over were the facts:
1)The columns of the echelon form of $A$ that do not contain leading row entries correspond to the "free variables" to $A{\bf x}={\bf 0}$. If the number of these columns is $r$, then the dimension of the null space is $r$ (again, if you know the dimension of the column space, you can see that the dimension of the null space must be the number of these columns from the rank-nullity theorem).
2) If you split up the general solution to $A{\bf x}={\bf 0}$ as done above, then these vectors will be independent (and span of course since you'll have $r$ of them).
Your system of equations can be converted into
$$\left\{ \begin{align}
x_1+x_3+x_5 = 0 \\
x_2+x_4 = 0
\end{align} \right.$$
so, $x_1=-x_3-x_5$ and $x_2=-x_4$. Hence, the vectors $(x_1,x_2,x_3,x_4,x_5)\in\mathbb R^5$ that satisfy the initial both equations also satisfy
$$\begin{align} (x_1,x_2,x_3,x_4,x_5)
&= (-x_3-x_5,-x_4,x_3,x_4,x_5) \\
&= x_3(-1,0,1,0,0)+x_4(0,-1,0,1,0)+x_5(-1,0,0,0,1)
\end{align}$$
that is, all of them can be written as a linear combinations of the vectors $(-1,0,1,0,0)$, $(0,-1,0,1,0)$ and $(-1,0,0,0,1)$. Thus,
$$\{(-1,0,1,0,0),(0,-1,0,1,0),(-1,0,0,0,1)\}$$
is a basis for $\textsf V$. Finally, apply the Gram-Schmidt process to this preceding set of vectors and we are done.
Best Answer
The null space of the matrix $$\begin{bmatrix} 1 & 1 & -1 & -1 & 0 & 0 \\ 1 & 1 & 0 & 0 & -1 & -1 \\ 0 & 0 & 1 & 1 & -1 & -1 \\ \end{bmatrix}$$ is, by definition, the set of vectors $\mathbf{x}=(x_1,x_2,x_3,x_4,x_5,x_6)$ such that \begin{align*} x_1 + x_2 - x_3 - x_4 &= 0 \\ x_1 + x_2 - x_5 - x_6 &= 0 \\ x_3 + x_4 - x_5 - x_6 &= 0 \end{align*} So, this is the vector space $V$ in question. The matrix has rank $2$, so the dimension of $V$ is $4$ by the Rank Nullity Theorem.
Thus, any $4$ linearly independent vectors in $V$ form a basis. A natural try is $$\{( 1, 1, 1, 1, 1, 1 ),(0, 0, 0, 0, 1, -1),( 0, 0, 1, -1, 0, 0 ),(1, -1, 0, 0, 0, 0)\}.$$ To prove that it is indeed a basis, we check that the matrix $$\begin{bmatrix} 1 & 1 & 1 & 1 & 1 & 1 \\ 0 & 0 & 0 & 0 & 1 & -1 \\ 0 & 0 & 1 & -1 & 0 & 0 \\ 1 & -1 & 0 & 0 & 0 & 0 \\ \end{bmatrix}$$ has rank $4$ (e.g. by performing Gaussian elimination).
The column space of the transpose of the above matrix, i.e. $$\begin{bmatrix} 1 & 0 & 0 & 1 \\ 1 & 0 & 0 & -1 \\ 1 & 0 & 1 & 0 \\ 1 & 0 & -1 & 0 \\ 1 & 1 & 0 & 0 \\ 1 & -1 & 0 & 0 \\ \end{bmatrix},$$ will be $V$ (or, at least, will be isomorphic to $V$: all the vectors will be transposed); see Wikipedia.