 Solution

False. Give $E=T{S}^2\to S^2$ a fibrewise area form. This is a rank-two symplectic vector bundle. A Lagrangian subbundle would be a real line subbundle $L\subset T{S}^2$. Real line bundles are classified by their first Stiefel–Whitney class in $H^1(B;\mathbb{Z}/2)$; since this group vanishes for $S^2$, $L$ would be trivial. A nonzero section of $L$ would therefore be a nowhere-vanishing tangent vector field on $S^2$. The Hairy ball theorem rules this out: every continuous tangent vector field on an even-dimensional sphere has a zero. Hence $T{S}^2$ does not even have one Lagrangian subbundle, let alone a splitting into two. Fibrewise existence of Lagrangian subspaces does not guarantee a globally continuous choice.