Yes: Denote by $E$ this family of subspaces. Choose $x\in X$ (with $\|x\|=1$) and $\phi_x\in Y$ with $\phi_x(x)=1$ and consider the weak star open set $U_x=\{\phi\in Y: \phi(x)\ne 0\}$. Then
$$
E|_{U_x} \ni (y,\phi) \mapsto (y -\phi_x(y).x,\phi)\in \ker(\phi_x)\times U_x
$$
is a trivializing vector bundle chart. The chart changes look like
$$
\ker(\phi_x)\times (U_x\cap U_y)\ni (z,\phi)\mapsto
\Big(z-\frac{\phi(z)}{\phi(x)}x,\phi\Big)\mapsto
\Big(z-\frac{\phi(z)}{\phi(x)}x - \big(\phi_y(z) - \frac{\phi(z)\phi_y(x)}{\phi(x)}\big)y,\phi\Big)
$$
$$
\in \ker(\phi_y)\times (U_x\cap U_y).
$$
So this is even vector bundle with rational transition functions.

# Edit:

Answers to the questions of the OP.

It is a sub vector bundle of the trivial bundle $X\times Y$ which fixes the topology,

The topology can also be induced by the vector bundle charts. Choosing an abstract linear isomorphism
from $\ker(\phi_x)$ to to a fixed Banach space (take $\ker(\phi_{x_0})$ for some fixed $x_0$) for every $x$, you may stabilize the fiber and give it a name.

finiteco-dimension are mutually isomorphic as normed vector spaces. In your case, $ \ker(\phi) $ has co-dimension $ 1 $ in $ X $ for each $ \phi \in Y $ and is closed in $ X $, hence complete. $\endgroup$ – Transcendental Feb 23 '15 at 18:19