The exercise says
Let $U$ a Banach subspace of a normed vector space $V$. Show that if $V/U$ is Banach then it is also $V$
Here $U$ have the norm induced from $V$ and the norm on $V/U$ is defined by $$ \|f+U\|:=\inf\{\|f+g\|:g\in U\} $$ I have a "proof" but probably it is wrong because I didn't used the fact that $U$ is Banach. Can someone show me my mistakes (if there are) please, or says if my proof is correct? In the case that it would be correct, we can drop then the condition that $U$ is Banach?
The proof below uses the following theorem
Let $(x_k)\subset V$ some sequence in a normed vector space $V$. Then $V$ is a Banach space if and only if $\sum_{k\geqslant 0}\|x_k\|<\infty\Rightarrow \sum_{k\geqslant 0}x_k\in V$.
My attempted proof of the exercise:
Let $(f_k)\subset V$ such that $\sum_{k\geqslant 0}\|f_k\|<\infty $, then we want to show that $\sum_{k\geqslant 0}f_k\in V$. Its easy to see that $\sum_{k\geqslant 0}\|f_k+U\|\leqslant \sum_{k\geqslant 0}\|f_k\|$ because $0\in U$, hence $\sum_{k\geqslant 0}f_k+U\in V/U$.
Then there is some $h\in V$ such that $\sum_{k\geqslant 0}f_k+U=h+U$, what imply that $\sum_{k\geqslant 0}f_k-h+U=U$ so $\sum_{k\geqslant 0}f_k-h\in U$ and consequently there is some $g\in U$ such that $\sum_{k\geqslant 0}f_k-h=g$. Hence $\sum_{k\geqslant 0}f_k=g+h$ and we are done.
Here is a (sketch of a) valid proof: let $(v_n)$ be Cauchy in $V$. Then $(v_n+U)$ is Cauchy in $V/U$. Let $v_n+U \to v+U$. Then there exists a sequence $(z_n)$ in $U$ such that $\|v_n-v-z_n\| \to 0$. Combine this with the fact that $(v_n)$ be Cauchy to show that $(z_n)$ is Cauchy. If $z_n \to z$ then $v_n \to v+z$.