Let $V$ be a vector space. If $U \leq V$, $Z \leq V$, $U \cap Z = 0$ and $Z \simeq V/U$, then $V = U \oplus Z$. Let $V$ be a vector space (possibly of infinite dimension).
I know that given a subspace $U$ of $V$, we can always write $V = U \oplus Z$, where $Z$ is some subspace of $V$ such that $Z \simeq V/U$.
Let $U$ and $Z$ be subspaces of $V$. I also know that, if $V = U \oplus Z$, then $Z \simeq V/U$.
Now I’m trying to prove the following statement: If $U \cap Z = 0$ and $Z \simeq V/U$, then $V = U \oplus Z$. Any help is appreciated.

The issue here is which map is giving you the isomorphism $Z\simeq V/U$. Let $\pi:V\to V/U$ denote the projection map.
Since every short exact sequence of vector spaces splits, there is a map $i:V/U\to V$ such that its composition with the projection above $i\circ\pi=id_{V/U}$ is the identity map. This gives you a way to identify $V/U$ with the subspace $im(i)=Z$ of $V$, i.e., an isomorphism $i:V/U\to Z\subset V$ which yields $V= U\oplus Z$. 
To detect a failure as in the example of the other answer you'd need to verify if the given isomorphism $f:Z\to V/U$ is compatible with the inclusion maps $j:Z\hookrightarrow V$ and $i: V/U\hookrightarrow V$: your claim holds if and only if $j=i\circ f$, i.e., both spaces are embedded as the same subspace of $V$.
