In Lax's Functional Analysis book:
The open sets in the weak topology are unions of finite intersections of sets of the form $\{x:a<l(x)<b\}$. Clearly, in an infinite-dimensional space the intersection of a finite number of sets of this form is unbounded.
I don't really see the "clearly" part. I may be missing something. Do we have to consider $\cap\ker f_i$ like the answer here?: Why unit open ball is open in norm topology, but not open in weak topology?
Sure, you can do it this way. If $y \neq 0$ is killed by all of the $f_i$, then if $x$ is in a basic weak neighborhood corresponding to the $f_i$, then $x+ty$ is also in this neighborhood for all $t$, since $f_i(x) = f_i(x+ty)$. In particular, the weak neighborhood will be unbounded.
Ultimately, I think the point is that the kernel of a linear functional is of codimension 1 in your space, and the intersection of $n$ kernels is therefore of codimension at most $n$, so there's a ton of wiggle-room inside of any basic neighborhood. Given any point in the neighborhood, you can translate it by anything in the intersection of these kernels without leaving the neighborhood. So the basic neighborhoods are huge.
