Lets say we have the following Matrix $$\left[ \begin{matrix} 2 & 0 & 0\\ 0 & 2 & 0\\ 0 & 0 & 0 \\ \end{matrix} \right]$$
Obviously a basis for this would be $$\begin{Bmatrix} \left[ \begin{matrix} 2\\ 0\\ 0\\ \end{matrix} \right] ,\left[ \begin{matrix} 0\\ 2\\ 0\\ \end{matrix} \right] \end{Bmatrix}$$ But would another valid basis be $$\begin{Bmatrix} \left[ \begin{matrix} 1\\ 0\\ 0\\ \end{matrix} \right] ,\left[ \begin{matrix} 0\\ 1\\ 0\\ \end{matrix} \right] \end{Bmatrix}$$
The answer to the question in the body of your post—is $$ \left\{\begin{bmatrix} 1 \\ 0 \\ 0 \end{bmatrix}, \begin{bmatrix} 0 \\ 1 \\ 0 \end{bmatrix}\right\} $$ a basis of the column space?—is yes! Remember the the column space of a matrix is spanned by the columns, and any spanning set contains a subset which is a basis. But of course there can be other bases for a subspace. The second set of vectors is linearly independent and has the same span as the first set, so is also a basis.
The answer to the question in the title of your post—must a basis for the column space consist of columns?—is no. That is the point of Abel's answer.
the answer is no to the question that should every basis of the column space of a matrix must have its columns in it. for example another basis for the column space is $\pmatrix{1 \\1\\0}, \pmatrix{1\\-1\\0}$. you can take any $ 2 \times 2 $ nonsingular matrix $P$ any basis $\{u_1,u_2\}$ then $\{Pu_1, Pu_2\}$ is another basis for the column space of $A.$
