I'm using the definitions from Rudins book Functional Analysis.
Suppose $X,Y$ are topological vector spaces, $\Lambda: X \to Y$ is continuous and linear and $\Lambda(X)$ is of the second category in Y.
The proposition is that $\Lambda$ is open (the actual proof requires X to be an F-space).
I came up with the following proof, which should be incorrect since I don't use the invariant complete metric on X, but I can't see where I go wrong.
Let $V$ be a neighborhood of $0$ in $X$. Choose a balanced neighbourhood of zero $U$ in $X$ such that $\bar{U} + \bar{U} \subset V$. (This is always possible, and used by Rudin, eg. in his proof of the Banach-Steinhaus theorem).
Now, $X = \cup _{n\in\mathbb{N}} n\bar{U}$, and since $\Lambda(X)$ is of the second category and $\Lambda(\cup _{n\in\mathbb{N}}n\bar{U}) \subset \cup_{n\in\mathbb{N}} n\Lambda(\bar{U})$ the latter union is of the second category, which implies that $\Lambda(\bar{U})$ is of the second category.
This implies $\exists y\in \Lambda(\bar{U})$ and an open neighbourhood $W$ of zero in $Y$ such that $y+W \subset \Lambda(\bar{U}) \implies W \subset\Lambda(\bar{U}) - y \subset \Lambda(\bar{U}) - \Lambda(\bar{U}) = \Lambda(\bar{U} - \bar{U}) \subset \Lambda(V)$.
Since $V$ was arbitrary, this proofs that $\Lambda$ is open at the origin, and therefore $\Lambda$ is open.
As pointed out by Daniel Fischer, the existence of $W$ is only guaranteed if $\Lambda(\bar{U})$ is closed, which in turn depends on $\Lambda$ being open..