Let $W,Z$ be Banach spaces. An (linear, bounded) operator $Q:W\rightarrow Z$ is said to be a quotient operator if it is surjective and $\Vert z\Vert=\inf\{\Vert w\Vert:Qw=z, w\in W\}$ (so $Z=W/\operatorname{ker}Q$).
I wanna show that an operator $Q:W\rightarrow Z$ with $\Vert Q\Vert=1$ is a quotient operator if for every $z\in Z$ and $\epsilon>0$, there is $w\in W$ such that $\Vert Qw-z\Vert<\epsilon$ and $\Vert w\Vert<(1+\epsilon)\Vert z\Vert$. Can you give me an idea?
Let $z\in Z$, with $\lVert z\rVert=1$ for simplicity. Pick any $\varepsilon>0$, and let $w_0\in W$ with $\lVert w_0\rVert<(1+\varepsilon)$ and $\lVert Qw_0-z_0\rVert<\varepsilon$. Let $z_1=z_0-Qw_0$, so $\lVert z_1\rVert<\varepsilon$.
Now, for $k=1$, $2$, …, assume that $z_k$ is given, and pick $w_k\in W$ with $\lVert w_k\rVert<2\lVert z_k\rVert$ and $\lVert Qw_k-z_k\rVert<2^{-k}\varepsilon$. Let $z_{k+1}=z_k-Qw_k$.
Note that $\lVert z_k\rVert<2^{1-k}\varepsilon$, and so $\lVert w_k\rVert<2^{2-k}\varepsilon$, for all $k\ge1$. We find therefore
$$z=z_0=Qw_0+z_1=Qw_0+Qw_1+z_2=\ldots=Q\sum_{k=0}^\infty w_k $$ where $$\Bigl\lVert\sum_{k=0}^\infty w_k\rVert\le \sum_{k=0}^\infty\lVert w_k\rVert < (1+\varepsilon)+\sum_{k=1}^\infty 2^{2-k}\varepsilon=1+5\varepsilon.$$ It follows that $Q$ is surjective, and $\lVert z\rVert\ge\inf\{\lVert w\rVert\in W\colon Qw=z\}$ for all $z\in Z$. The opposite inequality follows from $\lVert Q\rVert=1$.
I will prove this in the case that $W$ is reflexive. Consider a positive sequence $\{\epsilon_n\}_{n=1}^{\infty}$ such that $\epsilon_n$ decreases to $0$. Then, let $w_n$ be such that $\|Qw_n-z\| < \epsilon_n$ and $\|w_n\| < (1+\epsilon_n)\|z\|$. If $W$ is reflexive, then bounded sequences in $W$ have weakly convergent subsequences, so without loss of generality we let $w_n\rightharpoonup w$ for some $w\in W$. As $Q$ is weakly continuous, we have that $Qw_n\rightharpoonup Qw$. Then, as the norm is weakly lower semicontinuous in normed vector spaces, we have that $$\|Qw-z\|\leq \liminf_{n\to \infty} \|Qw_n-z\| = 0$$ which implies $Qw = z$. Therefore, $Q$ is surjective, as we can find such a $w$ for any $z\in Z$. Also, $$\|w\|\leq \liminf_{n\to \infty} \|w_n\|\leq \|z\|$$ Combined with the obvious fact that $Qw = z$ implies that $$\|z\| = \|Qw\|\leq \|w\|$$ we have that $\|w\| = \|z\|$. Clearly, we therefore cannot have a $w'$ such that $\|w'\| < \|w\|$ and $Qw' = z$, as $$\|Qw'\|\leq \|Q\|\|w'\| = \|w'\| < \|w\| = \|z\|$$ so we have that $\|z\| = \inf\{\|w\| : Qw = z, w\in W\}$, and $Q$ is a quotient operator.