I have been tasked with either proving that the edge space of every graph is generated by the cycles and cuts of the graph, or offering a counterexample. I think that $C_4$ is a counterexample, but I'm not quite sure, and I want to make sure I am at least understanding the concept of cycle space and edge space correctly.
Here is the counterexample. The cut space of $C_4$ is generated by cuts of the form $E(v)$, so cuts where the partition of $C_4$ is simply $(\{v\},C_4 \setminus \{v\})$. For $C_4$, the vectors of the edge space corresponding to these cuts are $(1,1,0,0), (0,1,1,0), (0,0,1,1), (1,0,0,1)$. Of course, $C_4$ is the only cycle in $C_4$, and the corresponding vector is $(1,1,1,1)$. My claim is that $(1,0,0,0)$ is not a linear combination of $(1,1,0,0), (0,1,1,0), (0,0,1,1), (1,0,0,1), (1,1,1,1)$, and consequently, $(1,0,0,0)$ is a vector in the edge space of $C_4$ that is not generated by the cuts and cycles of $C_4$.
We use some linear algebra: \begin{bmatrix} 1 & 1 & 0 & 0 & 1 & 1 \\ 0 & 1 & 1 & 0 & 1 & 0 \\ 0 & 0 & 1 & 1 & 1 & 0 \\ 1 & 0 & 0 & 1 & 1 & 0 \end{bmatrix} is the augmented matrix with the vectors generating the cut space and cycle space as well as $(1,0,0,0)$.
This thing row reduces to \begin{bmatrix} 1 & 1 & 0 & 0 & 1 & 1 \\ 0 & 1 & 1 & 0 & 1 & 0 \\ 0 & 0 & 1 & 1 & 1 & 0 \\ 0 & 0 & 0 & 0 & 0 & 1 \end{bmatrix} which implies that there is no solution. Thus, $(1,0,0,0)$ is a vector in the edge space that is not generated by the cuts and cycles of $C_4$.
It seems like a good argument to me, but it was easy enough that it seems too good to be true.
I think your proof is correct. I found the same question when I was reading Diestel's Graph Theory. In the hint to exercise 26, he mentioned that considering a cycle is useful.
However, your proof can perhaps be simplified a little. You only need to show $1,1,0,0),(0,1,1,0),(0,0,1,1),(1,0,0,1)$ cannot span $(1,0,0,0)$, as in $C_4$ the dimension of the edge space is $4$.
Hope this helps!