Consider $f:\mathbb{R}^{2} \rightarrow \mathbb{R}$ with $f(x,y)=\mathrm{e}^{x}\cos y$ then $\nabla (f)$ is nothing but $\mathrm{e}^{-z} :\mathbb{C} \rightarrow \mathbb{C}$, with image neither convex nor simply connected. This gives a negative answer to the second and the third part of your question.
Regarding the first part I do not know the complete answer. But I can say only the following: for every $V\subset\mathbb{R}^{n}$, $\nabla f[U]\cdot V$ is a connected subset of $\mathbb{R}$, because the partial derivatives satisfies Darboux theorem; hence they send open connected sets to connected subset of $\mathbb{R}$. Moreover, as a consequence of chain rule $\nabla f[U]\cdot V$ is a partial derivative. In fact there is no a hyper plane which separates $\nabla f[U]$.
So it is interesting to consider the following question:
Let $A$ be a subset of $R^{n}$, such that $A\cdot V$ is connected for all $V$, does this implies that $A$ is connected?
