 Solution

Because $Y$ is closed, it is a nowhere dense set exactly when its interior is empty. If a ball about $y\in Y$ lay inside $Y$, translating it by $-y$ would put a ball about $0$ inside $Y$. Every vector of $X$ has a small scalar multiple in that ball, and closure under scalar multiplication would force $Y=X$. This contradiction proves a proper closed linear subspace is nowhere dense.