Let $E$ be topological vector space Hausdorff, and $p: E \longrightarrow E$ be a continuous projection of $E$, that is, $p$ is linear and $p^2=p$, where $p^2=p$ means $$p(p(x))=p(x),\;\forall \; x \in E.$$ I want to prove that $p$ is open.
So, I think it's enough to prove that given $ x \in E $ then the set $ p(V) \in E$ is a neighborhood of $ p(x) \in E $, where $ V: = U_0 + x \subset E$ is a neighborhood of $ x $, with $ U_0 $ a neighborhood of $ 0 \in E $.
$p$ may not be open.
For example, let $E$ be R$^n$.
For all $x \in E$, let $p(x) = 0$.
Thus $p^2 = p$.
Now $E$ is open but $p(E) = {0}$ is not open.Consequently $p$ is not open.
The definition you gave for a projection is unusual.
It's clear that $p$ has a closed image in $E$, as $E$ is Hausdorff. So $p[E]$ is not open unless it equals $E$, by connectedness.
So $p$ can only be open as a map $E \to p[E]$. This would follow if $E$ were a Banach space (open mapping theorem), but I'm not sure it holds for any Hausdorff TVS $E$.
Put $M=P(E)$. It suffices to show that for any open neighborhood $U$ of $0\in E$, there is an open neighborhood $V\subset M$ of $0$ such that $V\subset P(U)$ (where the topology on $M$ is just the subspace topology).
Now fix an open neighborhood $U$ of $0$ in $E$, and let $V=U\cap M$. Then $V$ is an open neighborhood of $0$ in $M$. If $x\in V$, then in particular $x$ is in $U$, and $Px=x$ (since $x\in M$), so $x\in P(U)$ and thus $V\subset P(U)$.
