How do I show that given $M$ a closed subspace of a normed space $X$, and let $\pi$ be the canonical projection of X onto $X/M$. Prove that $\|\pi\| = 1$.
I figure I could use Riesz' lemma and set $\|\pi\| = 1$, that's as far as I got.
I could also use the fact that the canonical map is a contraction so I would have $\|\pi(x)\| \leq \|x\| \Rightarrow \frac{\|\pi(x)\|}{\|x\|} \leq 1$ and taking a supremum then we get the desired result. This doesn't seem as rigorous.
By definition of the norm on $X/M$, for each $x\in X$, $$\|\pi(x)\|=\inf\{\|x+y\|:y\in M\}\leq \|x+0\|=\|x\|,$$ hence $\|\pi\|\leq 1$.
For the other direction, we must assume that $M\neq X$; let $x$ be an element of $X\setminus M$. By closedness of $M$, $\|\pi(x)\|>0$, and thus for each $\varepsilon>0$, $\dfrac{\|\pi(x)\|}{1-\varepsilon}>\|\pi(x)\|=\inf\{\|x+y\|:y\in M\}$. This implies that there exists $y\in M$ such that $\|\pi(x)\|> \|x+y\|(1-\varepsilon)$. Thus $\|\pi(x+y)\|=\|\pi(x)\|> \|x+y\|(1-\varepsilon)$. Because $x+y\neq 0$, this implies that $\|\pi\|> 1-\varepsilon$.
