Column/Row Space check I have the following matrix: 
\begin{bmatrix}
    1 & 2 & 0 & 1 & 0\\
    3 & 6& 1 & 6 & 1\\
    2 & 4 & -1 & -1 & -1\\
   4 & 8  & 0 & 4 & 0\\
  \end{bmatrix}
and I reduce it to 
\begin{bmatrix}
    1 & 2 & 0 & 1 & 0\\
    0 & 0& 1 & 3 & 1\\
    0 & 0 & 0 & 0 & 0\\
   0 & 0  & 0 & 0& 0\\
  \end{bmatrix}
I have some confusion on what the column space of the matrix should be. Am I correct in saying that the column space is: span $ \left \{ (1,0,0,0)^T,\; (0,1,0,0)^T \right \}$. The Row space would be the span of the basis vectors of the row space: $\left \{ (1,2,0,1,0),\; (0,0,1,3,1) \right \}$.
Am I correct?
 A: Only the row space is preserved
Regarding the column space, row reduction changes that drastically so although $A\sim H$ we cannot say that $A$ and $H$ have the same column space. But row operations does not change the row space since the rows of $H$ are still in the span of the rows of $A$. Also the row space of $A$ cannot contain fewer vectors than the nonzero rows of $H$ since these are clearly independent (ask if it is unclear). And since row operations are invertible the nonzero rows of $H$ have all nonzero rows of $A$ in their span.
The column space is found indirectly
The column space is spanned by the columns in the original matrix $A$ corresponding to the pivots of the row reduced echelon form $H$. In your case that is $\{(1,2,3,4)^\mathrm T,(0,1,-1,0)^\mathrm T\}$.

Explanation regarding the column space: Your calculations show that
$$
A=\begin{bmatrix}
    1 & 2 & 0 & 1 & 0\\
    3 & 6& 1 & 6 & 1\\
    2 & 4 & -1 & -1 & -1\\
   4 & 8  & 0 & 4 & 0\\
  \end{bmatrix}
\sim
\begin{bmatrix}
    1 & 2 & 0 & 1 & 0\\
    0 & 0& 1 & 3 & 1\\
    0 & 0 & 0 & 0 & 0\\
   0 & 0  & 0 & 0& 0\\
  \end{bmatrix}
=H
$$
Therefore $Ax=0$ is equivalent to $Hx=0$ (why is that?). In particular $x$ equal to $(-2,1,0,0,0)^\mathrm T$, $(-1,0,-3,1,0)^\mathrm T$ and $(0,0,-1,0,1)^\mathrm T\}$ solves $Hx=0$ and therefore $Ax=0$ showing (if $a_i$ denotes the $i$-th column vector of $A$) that
$$
\begin{align}
-2a_1+a_2&=0&&\iff&&&a_2&=2a_1\\
-a_1-3a_3+a_4&=0&&\iff&&&a_4&=a_1+3a_3\\
-a_3+a_5&=0&&\iff&&&a_5&=a_3
\end{align}
$$
This shows that the column vectors $a_2,a_4$ and $a_5$ can be expressed via $a_1$ and $a_3$ so they are already in $\operatorname{span}(a_1,a_3)$.
