$X+Y$ is closed $\Leftrightarrow$ $\|x\|\leq c\|x+y\|$ for all $x\in X$ and all $y \in Y$. The problem says

Let $(Z,\|\cdot\|)$ be a Banach space. Let $X$ and $Y$ be two closed subspaces of $Z$ such that $X\cap Y=\{0\}$. Prove that $X+Y$ is closed if, and only if, there exists $c\geq 0$ such that
  $$\|x\|\leq c\|x+y\|,\quad \forall\ x\in X,\quad\forall \ y \in Y.\tag{1}$$

My attempt: Suppose that condition $(1)$ holds. Let $(w_n)$ be a Cauchy sequence in $X+Y$. Then, for all $n$, we have $w_n=x_n+y_n$ with $x_n\in X$ and $y_n\in Y$. By $(1)$ we have
$$\|x_n-x_m\|\leq c\|w_n-w_m\|,\quad \forall \ n,m\in\mathbb{N}.$$
It follows that $(x_n)$ and $(y_n)$ are Cauchy sequences. Since $X$ and $Y$ are complete, $(x_n)$ converges to some $x\in X$ and $(y_n)$ converges to some $y\in Y$. Taking $z=x+y$ we see $(z_n)$ converges to $z\in X+Y$. So, $X+Y$ is complete and thus closed.
1) In my argumment, the hypothesis $X\cap Y=\{0\}$ wasn't used. It's correct? There are any alternative approach?
2) Please, hints for the converse.
Thanks.
 A: A suggestion for the converse: since $X \cap Y = 0$, then you have a direct sum. This means that any vector $v \in X \oplus Y$ has a unique decomposition as $x+y$, with $x \in X$ and $y \in Y$. Define the operator $U : X \oplus Y \rightarrow X$, $U(v)=x$ (the projection onto the first summand). Choose $c= \Vert U\Vert$ (prove that it is finite).
A: For the inverse,
consider
$X+Y$ is closed, but (1) is not sure.
This means
$\forall c\geqslant 0$ we can find a $x\in X$ and $y\in Y$ such that,
$\Vert x\Vert>c\Vert x+y\Vert $
divided by $\Vert x\Vert$ we can assume that $x$ has norm 1.
then for any $\frac{1}{c}>0$ we can find 
$\Vert x+y\Vert<\frac{1}{c}$,
now
$\vert\Vert x\Vert-\Vert y\Vert\vert\leqslant\Vert x+y\Vert$.
So,
$\Vert y\Vert$ can be also assumed to be bounded.
notice,
$\Vert x+y\Vert$+$\inf_{x'\in X}\Vert y-x'\Vert\geqslant\inf_{x'\in X}\Vert x+y+y-x'\Vert$=$\inf_{x'\in X}\Vert 2y-x'\Vert$=$d(2y,X)$.
also,
$\inf_{x'\in X}\Vert 2y-x'\Vert$=$2\inf_{x'\in X}\Vert y-x'\Vert$=$2d(y,X)$
which means,
$2d(y,X)\leqslant d(y,X)+\Vert x+y\Vert$
so,
$d(y,X)$ can be any small.
denote $Y_M^m$=$\{y\in Y\vert m\leqslant\Vert y\Vert\leqslant M\}$, since we assume $y$ to be bounded we can find such $M$ that fits our situation.
$d(Y_M^m,X)$ can not be $0$ by the closeness of those sets.
$d(y,X)$ can be any small means we can choose $y\notin Y_M^m$ but the norm is bounded by $M$, therefore by the arbitrary nature of smallness of $d(y,X)$ we have to conclude that $y=0$ but then $x$ will have to be $0$, not as we required to be norm $1$.
Ok, in the whole discuss we can choose $\frac{1}{c}$ to be $\frac{1}{2^n}$ and denote the sequence $x_n,y_n$ and do the above argument. 
