Passage in proof existence of Symplectic structure on Symplectic fibration I need help in understanding the last passage in Thurston’s proof of the existence of a compatible Symplectic form on a Symplectic fibration $M\to B$ With Symplectic base space $(B,\beta)$. 
More precisely the setting is the following. Let $i_b\colon M_b\to M$ be the fiber inclusion, and assume every fiber is a Symplectic manifold with given fiberwise Symplectic forms $\sigma_b \in \Omega^2(M_b)$.

Assume there exists a closed $2$-form $\alpha \in \Omega^2(M)$ with $$i_b^*[\alpha]=[\sigma_b] \ \ \ \ (*)$$ we can find a compatible Symplectic structure by adding sufficiently positive multiple of $\pi^*\beta$ to $\alpha$.

I’m following the proof in McDuff Salamon “introduction to Symplectic geometry” (Thm 6.3 page 199) where they carefully tweak $\alpha$ to have $(*)$ valid at the cochain level and then claim that by adding suitably big multiples of $\beta$ we have that $$\alpha + K\pi^*\beta$$ is a Symplectic form on $TM$.
Can someone explain to me why such $K>>0$ must exist?
 A: This is a bit late answer to the question. Perhaps you already know why such $K$ exists.
Since $\omega:= \alpha + K\pi^*\beta$ is closed, it remains to show $\omega$ is nondegenerate. Given a point $x \in M$, let $V:= \mathrm{Ker} \ d\pi_x$ be the tangent space of the fiber. Since $V$ is a symplectic subspace of $T_xM$, we have a direct sum decomposition

$T_x M = V \oplus H, \quad$ where $H = \{v \in T_xM \mid \alpha(v,w) = 0 \text{ for all } w \in V\}$.

Write $X, Y \in T_xM$ as $X = X_V + X_H$ and $Y = Y_V + Y_H$ with respect to the above decomposition. Then,

\begin{align}\tag{*}
\omega(X,Y) &= \alpha(X,Y) + K\pi^*\beta(X,Y) \\
&= \alpha(X_V + X_H, Y_V + Y_H) + K\beta(X_H, Y_H) \\
&= \alpha(X_V, Y_V) + \alpha(X_H, Y_H) + K\beta(X_H, Y_H).
\end{align}

Given nonzero $X$, we need to find $Y$ such that the above expression is not zero. When $X_H=0$, this follows from the fact that $\alpha$ is symplectic along the fiber. When $X_H \neq 0$, choose $Y$ so that $\beta(X_H, Y_H) \neq 0$. Since the values of the first two terms are fixed, we can find $K>0$ so that the whole value is nonzero.
So far $K$ depends on the choice of $x \in M$ and also $X, Y \in T_x M$. It remains to show why $K$ can be uniformly chosen. Since the expression (*) is bilinear, after fixing a Riemannian metric, we may assume that $X$ and $Y$ have unit length. Since the unit tangent bundle is compact (we need an assumption that $M$ is compact) and $\omega(X,Y) \neq 0$ is an open condition, we can find finite $K>>0$ so that $\omega$ is nondegenerate.
