Question about statement of Rank Theorem in Rudin 
Theorem Suppose $m,n,r$ are nonnegative integers, $m\ge r, n\ge r$, $F$ is a $C^1$ mapping of an open set $E\subset \mathbb{R}^n$ into $\mathbb{R}^m$, and $F'(x)$ has rank $r$ for every $x\in E$. Fix $a\in E$, put $A = F'(a)$, let $Y_1$ be the range of $A$, and let $P$ be a projection in $\mathbb{R}^m$ whose range is $Y_1$ and let $Y_2$ be the kernel of $P$. Then there are open sets $U$ and $V$ in $\mathbb{R}^n$, with $a\in U\subset E$, and there is a 1-1 $C^1$ mapping $H$ of $V$ onto $U$ (whose inverse is also of $C^1$) such that $$F(H(x)) = Ax+\phi(Ax)\;\;\;\;(x\in V)$$where $\phi$ is a $C^1$ mapping of the open set $A(V)\subset Y_1$ into $Y_2$.

Two questions: 1. Why is $A(V)$ a open set?
2. What does this theorem really say? I lost intuition due to its complicated formulation.
 A: Yeah, I think this is a terrible way of looking at the rank theorem. A good way to understand the rank theorem is to consider its consequences, i.e. the inverse and implicit function theorems. I'll try to give a better explanation below. To answer 1, $A(V)$ is open because $V$ is open and $A$ is linear, and non-degenerate linear maps are open. For instance, in $d = 1$ and $A \subset \mathbb{R}$, this boils down to 
$$A \text{ open } \Rightarrow \{cx : x \in A\} \text{ open }, c \neq 0.$$
Hopefully if you're looking at the rank theorem you know how to generalise this to vector spaces and norms... 
As for what the rank theorem is actually saying, let me ask you this question: what is the dimension of the sphere $\mathbb{S}^2 \subset \mathbb{R}^3$? Intuitively, it is 2. Why? Well, save the point at infinity, I have a bijection between points on a plane and points on the sphere (stereographic projection), and a plane is two-dimensional. The rank theorem allows you to say the dimension of the sphere is 2 without knowing about this map. It characterises the dimension of the surface in terms of the rank of the tangent plane at each point. The tangent plane is spanned by 2 linearly independent vectors at every point of the sphere. 
You can think about this in many ways -- here are two. Let $E = (0,\pi) \times (0, 2\pi)$ and $F(\phi,\theta) = (\sin\phi\cos\theta,\sin\phi\sin\theta,\cos\phi)$. If you compute $F'$, you'll notice that the rank of $F$ is 2 at every point (save degeneracies), so by the rank theorem, for any point on the sphere there is a small patch around it (a neighbourhood) and a $C^1$ map which takes that curved patch to a flat patch $\subset \mathbb{R}^2$. 
Another way to do this is to set $$G(x,y,z) = x^2 + y^2 + z^2,$$ $G: \mathbb{R}^3 \rightarrow \mathbb{R}^1$ so that $\nabla G = (2x,2y,2z)$. The rank is 1 for $(x,y,z) \neq 0$, and $\nabla G$ is the normal to the tangent plane, which by the rank theorem always has rank 1. Thus the tangent plane to the sphere always has rank $3 - 1 = 2$ (the 'codimension' of the normal vector in $\mathbb{R}^3$). 
A good thinking point is to go back like 150 years. People were really curious about this question: if I have a surface described by a level set,
$$F(x,y,z) = 0,$$
can I always write it explicitly as a graph,
$$z = f(x,y)?$$
In the case of a sphere, sure:
$$x^2 + y^2 + z^2 = 1 \Rightarrow z = \pm \sqrt{x^2 + y^2},$$
but you can certainly cook up examples where this is a lot more difficult (try cubics). The implicit function theorem, a consequence of the rank theorem, answers this question in the affirmative, contingent on properties of the rank of the derivative. This makes sense if you think about the one-dimensional case: you can invert a function $y = f(x)$ at a point $x_0$ if $f'(x_0) \neq 0$ -- in this case the rank is 1 and the dimension is 1, but it's the rank theorem (or inverse function theorem) at work.
EDIT: As a way of remembering the statement, just think about linear maps. If 
$$A = \begin{pmatrix} \cdot & \cdot & \cdot & \cdot \\ \cdot & \cdot & \cdot & \cdot \end{pmatrix},$$
$A : \mathbb{R}^4 \rightarrow \mathbb{R}^2$, and you can get at most two dimensions back -- the other two dimensions must get killed. This is the case $m = 2$, $n = 4$, and the rank can be at most $2$. The image can be the whole space. If 
$$A = \begin{pmatrix} \cdot & \cdot & \cdot \\ \cdot & \cdot & \cdot \\ \cdot & \cdot & \cdot \\ \cdot & \cdot & \cdot \end{pmatrix},$$
$A : \mathbb{R}^3 \rightarrow \mathbb{R}^4$ and you can get all your dimensions back, but you can't span the whole target space. 
