Unused hypothesis in projection proof 
Let $E$ be a Banach Space and $M$ and $N$ be closed subspaces. If every $x \in E$ is written uniquely in the form $x=p+q$ for $p \in M$ and $q \in N$ then defining $P(x) = p$ and $Q(x) = q$, prove that $P,Q$ are linear and continuous operators and that they are projections.

This exercise is listed under "Closed graph theorem" and we saw projections as linear continuous and idempotent operators (on normed spaces), I don't see how the Banach hypothesis is needed or why $M$ and $N$ are listed as closed (they indeed end up being as a consequence of the projection and complementation). I shall prove it for $P$ and the proof is analogous to $Q$, $P$ is trivially well defined and linear through the uniqueness of the decomposition, and for continuity:
$$ \|P(x)\| = \|P(x)+Q(x)-Q(x)\| = \|x-Q(x)\| \le \|x\|+\|Q(x)\| \le \|x\|$$
Therefore , $\|P\|\le 1$ and therefore it is continuous, it is a projection since,through uniqueness $P(p) = p$ and then $P(P(x)) = P(p) = p = P(x)$  for $x \in E$ (letting $x=p+q$), furthermore $P(E)=M$.
I can't see anything wrong with my reasoning so I assume the exercise just has extra hypothesis for the sake of it, but it's a lot simpler than most others in this section so I'm a little suspicious.
 A: All assumption are essential. The proof that $P$ is continuous goes like this: assume that $x_n \to x \in E$ and that $Px_n \to p \in E$. We wish to prove $p = Px$.
Since $M$ is closed and $(Px_n)_n$ is a convergent sequence contained in $M$, it follows $p \in M$. We wish to prove that $$x = p + (x-p)$$
is the unique decomposition with $p \in M$ and $x-p \in N$. Clearly it remains to show that $x-p \in N$. If we denote $x_n = p_n + q_n$ its unique decomposition, we have $p_n \to p$ and hence
$$q_n = x_n - p_n \to x-p.$$
Since $(q_n)_n$ is a convergent sequence contained in a closed subspace $N$, it follows that $x-p \in N$. It follows $p = Px$. Since $E$ is a Banach space, by the Closed Graph Theorem it follows that $P$ is continuous.
Now $Q = I-P$ is continuous as well.
Now to demonstrate that the subspaces have to be closed. Let $E$ be an infinite-dimensional Banach space and let $f : E \to \Bbb{F}$ be an unbounded linear functional. Pick $x_0 \in E$ such that $f(x_0) = 1$. Then set $M := \Bbb{F}x_0$ and $N := \ker f$. Then every $x \in M$ can be uniquely written as $$x = \underbrace{f(x)x_0}_{ \in M} + \underbrace{x-f(x)x_0}_{\in N}$$
but neither projection is continuous. The problem is that $N$ is not a closed subspace of $E$.
Now we shall demonstrate that $E$ has to be Banach, even if $M$ and $N$ are closed. Consider $c_{00}$, the space of all finitely supported sequences, and closed subspaces
$$M = \{(x_n)_n \in c_{00} : x_{2n}=0, \forall n \in \Bbb{N}\}, \quad N = \{(x_n)_n \in c_{00} : x_{2n-1}+na_{2n}=0, \forall n \in \Bbb{N}\}.$$
Then every $(x_n)_n \in c_{00}$ can be uniquely written as $$(x_n)_n = \underbrace{(x_1+x_2, 0, x_3+2x_4, 0, x_5+3x_6, 0, \ldots)}_{\in M} + \underbrace{(-x_2,x_2,-2x_4,x_4,-3x_6,x_6, \ldots)}_{\in N}$$
but the projections are clearly unbounded.
