Sufficient condition for the sum of two closed subspaces of a Banach space to be closed. Let $X$ be a Banach space and $Y$ and $Z$ its closed subspaces such that $Y \cap Z = \{0\},$ and $$k = \inf \{||y-z|| \ | \ y\in Y, z \in Z, ||y||= ||z||= 1 \}.$$
I need to show that if $k > 0$ then $Y+Z$ is closed. I managed to show the other way around, however here I am stuck. If anyone has any idea how to proceed I would be really greatfull.
 A: Let us first prove that there is a constant $C\geq 0$, such that
$$
  \|y\|\leq C\|y-z\|,  \quad\forall y\in  Y, \quad\forall z\in  Z.
  \tag 1
  $$
Arguing by contradiction we could otherwise find sequences $\{y_n\}_n$ and $\{z_n\}_n$ in $Y$ and $Z$,  respectively,  such that
$\|y_n\|=1$,  while $\|y_n-z_n\|\to 0$.
Observe  that
$$
  \big|\|z_n\|-1\big| =    \big|\|z_n\|-\|y_n\|\big| \leq   \|z_n-y_n\| \to 0,
  $$
so, defining $z'_n = z_n/\|z_n\|$, one has that $\|z_n-z'_n\|\to 0$.
Since $\|y_n\|=\|z'_n\|=1$, we have by hypothesis that
$$
  k\leq \|y_n-z'_n\|\leq   \|y_n-z_n\|+   \|z_n-z'_n\| \to 0,
  $$
a contradiction, hence proving (1)
Replacing $z$ by $-z$ in (1) one also gets
$$
  \|y\|\leq C\|y+z\|,  \quad\forall y\in  Y, \quad\forall z\in  Z.
  \tag 2
  $$
Similarly one proves the existence of a constant $D\geq 0$, such that
$$
  \|z\|\leq D\|y+z\|,  \quad\forall y\in  Y, \quad\forall z\in  Z.
  \tag 3
  $$
Now, to prove that $Y+Z$ is closed, let $x$ be a point in the closure, so we may write
$$
  x=\lim_{n\to \infty }y_n+z_n,
  $$
with $\{y_n\}_n\subseteq Y$, and $\{z_n\}_n\subseteq Z$.
Clearly $\{y_n+z_n\}_n$ is a Cauchy sequence,  whence so are $\{y_n\}_n$ and $\{z_n\}_n$  by (2) and (3),  so we may define
$$
  y=\lim_{n\to \infty }y_n,  \quad\text{and}\quad z=\lim_{n\to \infty }z_n.
  $$
It then follows that
$$
  x = \lim_{n\to \infty }y_n+z_n = \lim_{n\to \infty }y_n + \lim_{n\to \infty }z_n =  y+z \in  Y+Z.
  $$
