Let $(M,g)$ a $n$ dimensional Riemannian manifold, $U \subset M$ an open subset and $E \in \mathfrak{X}(U)$ a unit parallel vector field (i.e.$\nabla{E}\equiv0$ and $g(E,E)\equiv1$).
Now consider the smooth $(n-1)$-rank distribution $D$ on $U$ defined by $D_m=\langle E_m \rangle^\perp <T_mU$ for all $m \in U$. It's easy to show that $D$ is involutive using $\nabla{E}\equiv0$, so by Frobenius' theorem $D$ is integrable.
Fix $m \in U$, and let $F$ the integral manifold containing $m$. Using the flowout $\phi$ of $E$ respect to the submanifold $F$ we have a diffeomorphism $$\phi \colon F\times I \to M $$ around $(m,0)$ so, given $(x_1,\dots,x_{n-1})$ coordinates on $F$ around $m$, we have coordinates $(x_1,\dots,x_{n-1},t)$ on a neighborhood $V \subset M$ of $m$.
I want to show that in $V$ we have a Riemannian product. I can see that $g_p$ is the product metric of the metric of $F$ and the metric of an euclidean line for all $p \in F$ but what about the points in $U\setminus F$?
As pointed out by Romulus $E$ is also a Killing vector field and an important property is that the flow of a Killing vector field is an isometry so using this it follows easily that the metric in $V$ is a product.