Proof verification that $(V/U)/(W/U)\cong (V/W)$ I have already shown that if $A:V\rightarrow X$ is a linear map between two vectorspaces then 
$V/\ker A\cong \text{Im}A$
To prove the Statement I have to find a map $V/U\rightarrow V/W$ which is linear, surjective and where the $\ker$ is $W/U$. I know that $W$ is a subspace of $V$ and that $U$ is subspace of $W$
I have Chosen then map $\phi:V/U\rightarrow V/W$ with $\phi(v+U)=v+W$.
The map is surjective because if $x\in V/W$ then there exists a $v\in V$ such that $x=v+W$ then $\phi(v+U)=x$
The map is also linear because $\phi((v+z)+U)=(v+z)+W=v+W+z+W=\phi(v+U)+\phi(z+U)$ and
$\phi(\lambda v+U)=\lambda v+W=\lambda (v+W)=\lambda (\phi(v+U))$
And also $\ker\phi=W/U$ Suppose $x\notin W/U$ then $x=v+U$ and $v\notin W$ and $\phi(x)\neq W\Rightarrow x\notin \ker\phi$. This means by contrapsoition $\ker \phi\subseteq W/U$. On the other hand $x\in W/U\Rightarrow x\in \ker \phi$. Then $W/U\subseteq\ker \phi\Longrightarrow W/U=\ker\phi$
Is this proof right?
 A: To prove well definition, note that $v+U = z+U$ if and only if $v-z \in U$. But by assumption, $U \subseteq W \subseteq V$. Hence, $v -z \in W$, and thus $v+W = z +W$.
Your proof is right, but I have a couple of nitpicks lol. To prove linearity, you should have another equal sign in the beginning:
\begin{equation}
\phi[(v+U) + (z+U)] := \phi[(v+z) + U] = \dots
\end{equation}
the first equal sign being by definition of addition in the quotient space. Similarly for scalar multiplication:
\begin{equation}
\phi[\lambda \cdot (v+U)] := \phi[(\lambda v) + U] = \dots
\end{equation}
I hope you see why if you were to strictly follow the definition of linearity, you'd have to start from here.
Also, as a rule of thumb, try to avoid contrapositive/contradiction whenever a direct proof is just as quick (it's fewer negations for the reader to verify).  In this specific example, you could have given a direct proof that $\ker \phi \subseteq W \setminus U$. (I know there are cases where contrapositive/contradiction is easier to read/quicker... but this IMO isn't one of them)
