Is an NVS complete iff it is non meagre?

Each infinite dimensional Banach space $X$ contains a dense hyperplane which is of second category in itself:

Let $\{b_j: j\in J\}$ be a Hamel base of $X$, thus each $x \in X$ has a representation $$ x= \sum_{j \in J} \varphi_j(x)b_j $$ with $\varphi_j(x) \not= 0$ for at most finitely many $j \in J$. For each $j \in J$ let $Y_j$ be the kernel of the linear functional $\varphi_j$. Assume by contradiction that $Y_j$ is of first category in $X$ for infinitely many $j \in J$. Let $I \subseteq J$ be a countable infinite subset such that $Y_j$ is of first category $(j \in I)$. Then for each $x \in X$ there is some $j \in I$ with $\varphi_j(x)=0$. So $$ X=\bigcup_{j \in I} Y_j $$ is of first category, a contradiction. Thus there are (many) $Y_j$ such that $Y_j$ is of second category in $X$. Such an $Y_j$ is then also of second category in itself. Moreover it can't be closed (otherwise it would be a closed strict subspace of $X$, hence of first category). By the same reasoning it is dense in $X$ and therefore it is not complete.