Smooth fibration is a submersion

Let $x$ be a point of $B$, consider a chart $f:U\simeq I^n\rightarrow B$ whose domain contains $x$ and $f(0)=x$. Write $Y=\{x\}$ and consider a point $z\in p^{-1}(x)$. You can define $\tilde f:Y\times \{0\}\rightarrow E$ by $\tilde f(x,0)=z$. Let $F:\{x\}\times I^n\rightarrow B$ defined by $F(x,y)=f(x)$. We have $p\circ \tilde f=F\circ i$. The homotopy lifting property implies the existence of $H:\{x\}\times U\rightarrow E$ such that $p\circ H=f$. Since the tangent map $df_0:T_0\mathbb{R}^n\rightarrow T_xB$ is bijective, and $df_0=dp_z\circ dH_{(0,x)}$, we deduce that $dp_z$ is surjective, and henceforth $p$ is a submersion.