When finding the basis of a subspace and having a free variable, would the dimension be just one? 
So I am not sure if this is the correct way and answer to this problem but here is my solution


The problem:
\begin{equation*}
{\bigg( x,y,z \bigg)| x+y=0 \space and \space x-2z=0}
\end{equation*}
Solution:
$$ 
x  \ -2z=0\\
    x+y   =0\\x=2z\\y=-2z\\So,\\ \bigg(x,y,z\bigg)=\bigg(2,-2,1)z\\ basis :\\<(2,-2,1)>\\dimension \space is \space 1
$$


So is my solution correct and would it be a dimension of 1?


Indeed your work is correct. Another way of looking at it would be geometrically. The set of points in question you're looking at is $$ \{(x,y,z): x+y=0 \text{ and } x-2z=0\}$$ can be thought of as the intersection of two planes. In this case since the planes are not parallel, the intersection is an object of dimension 1, a line, whose direction vector you've found in your solution.
