Can someone illustrate the definition of manifold with a simple example? In my text the definition of a differential manifold is given as follows:

A subset $M$ of $\mathbb{R}^n$ is a $k$ dimensional manifold if
  $\forall x \in M$ there are open subsets $U$ and $V$ of $\mathbb{R}^n$
  with $x \in U$ and diffeomorphism $f$ from $U$ to $V$ 
such that
$f(U ∩ M) = \{y \in V: y^{k+1} =...=y^n = 0\}$ 
Thus a point $y$ in $im(f)$ has representation $y = (y^1(x), y^2(x),...,y^k(x), 0 ...0)$

There are too many sets going on at the same time I am totally lost. Can someone motivate this definition through a simple example? i.e. showing that a line is a 1D manifold in 1D, 2D spaces
 A: a good example is the 2-sphere $\Bbb{S}^2$ which is described in $\Bbb{R}^3$ by the equation:
$$
x^2+y^2+z^2=1
$$
this is a 2-dimensional surface but you cannot give a 2-parameter system of co-ordinates which works for the whole sphere (i.e. globally). this situation became familiar to 18-th century mathematicians as the projection problem for maps of the entire surface of the spheroidal Earth. for example the coordinates $(x,y)$  work OK for the northern hemisphere or for the southern hemisphere, but if this system is applied to the whole sphere any point $(x,y,z)$ with $z \ne 0$ has the same coordinates as the point $(x,y,-z)$ 
coordinates must be unique. 
so we do the best we can, and restore uniqueness at the expense of global coverage. basically we cover $\Bbb{S}^2$ with a family of open sets $U^{\alpha}$ and give a unique set of coordinates for each $U^{\alpha}$. by set of coordinates we mean a map $\phi_{\alpha}:U^{\alpha} \to U_{\alpha}$ where $U_{\alpha} $ is an open set in $\Bbb{R}^2$ and the map $\phi_{\alpha}$ is a bi-unique map. usually the $\phi_{\alpha}$ are required to satisfy a smoothness condition (in both directions), such as being twice-differentiable, or infinitely differentiable, or (in the case of complex manifolds) being a holomorphic function. in the slightly simplified notation employed here, it makes sense to use the symbol $\phi^{\alpha}$ to signify the inverse map $\phi_{\alpha}^{-1}$.
this use of upper and lower indices will make intuitive sense if you think of the manifold as lying over $\Bbb{R}^2$
note that $\phi^{\alpha} \circ \phi_{\alpha}$ is the identity map on  the open set $U^{\alpha}$ in the manifold whereas $\phi_{\alpha} \circ \phi^{\alpha}$ is the identity on the corresponding open subset $U_{\alpha}=\phi_\alpha(U^\alpha)$ of $\Bbb{R}^2$
this use of indices makes sense if you think of an open subset of the manifold as above its coordinate image in $\Bbb{R}^2$
a pair $(U^{\alpha}, \phi_\alpha)$ is called a co-ordinate chart and, of course, $\phi_\alpha$ takes the form of an ordered pair of real-valued functions $(\phi_{\alpha,x},\phi_{\alpha,y)}$, so that for any point $p$ in the manifold which lies in the chart domain $U^\alpha$ we have the coordinates $x=\phi_{\alpha,x}(p)$ and $y=\phi_{\alpha,y}(p)$
we manage the global coverage by ensuring that we have enough charts that their domains form an open cover of the manifold. such a collection of charts is called an atlas (which continues the historically appropriate cartographic metaphor).
unless the manifold has more than 1 connected component and each component can be dealt with by a single chart (this occurs, for example, in a manifold consisting of a set of two or more mutually disjoint open line segments) we can only restore the global coverage by having some of the chart domains overlap. where overlap occurs we have to have in induced smooth map between two subsets of $\Bbb{R}^2$. (they may or may not overlap). this requires that both the following composite maps must be smooth, bi-unique maps between subsets of $\Bbb{R}^2$.
$$
\phi_\beta \circ \phi^\alpha: \phi_\alpha(U^\alpha \cap U^\beta) \to \phi_\beta(U^\alpha \cap U^\beta)
$$
and
$$
\phi_\alpha \circ \phi^\beta: \phi_\beta(U^\alpha \cap U^\beta) \to \phi_\alpha(U^\alpha \cap U^\beta)
$$
for the sphere we could almost get by with two charts as defined earlier for opposite hemispheres. however the equator is excluded. this is no mere technical problem, and thinking about it, in terms of the two conditions just described clarifies the reason why chart domains must be open subsets.
a simple atlas for $\Bbb{S}^2$ is obtained using six charts whse domains are, respectively, the three pairs of open hemispheres sliced off by the planes $x=0,y=0,z=0$.
each pair has an equator that cannot be mapped. if we have two pairs then the two poles - where the corresponding equators meet - still pose a problem. but when we have all three pairs of charts there is no point left unaccounted for. 
although the two-dimensional layout of a city could be shown with a single chart,  in fact, for convenience, an atlas is used even in this case where it is not, as in the case of the sphere, obligatory. the $A-Z$ books of maps of towns do this, and you will notice that the chart domains overlap at the edges, with indicator numbers referring to the appropriate continuation pages. these represent the (linear tranlation) the maps $\phi_\alpha \circ \phi^\beta$.
the idea of a coordinate atlas is simple enough. what takes time and effort is to  adjust your thinking to using chart maps and their inverses to transport yourself in either direction between a neighbourhood on the manifold and a neighbourhood in $\Bbb{R}^2$. for it is in $\Bbb{R}^2$ that we can apply the powerful machinery of the calculus.
a bonus comes when you realize it is possible to define a manifold in terms of the transformations between the charts of an atlas even if we are not given a global embedding into higher dimensional real or complex orthogonal space.
A: In simple words, $M$ is a $k$-dimensional manifold in $\mathbb{R}^n$ means that for all $x\in M$, there exists $U$ a neighborhood of $x$ such that $U\cap M$ is diffeomorphic to an open subset of $\mathbb{R}^k$.Such a local diffeomorphism has to exist for every point of $M$.
Example : around every point of a circle, you can find an open neighborhood that can be mapped (with a diffeomorphism) to an open subset of $\mathbb{R}$. That's why a circle is a 1-dimensionnel manifold.
Intuitively, it is the "real dimension" of the object.
A: $\newcommand{\Reals}{\mathbf{R}}$Let $k$ and $n$ be non-negative integers with $k \leq n$. The simplest example of a $k$-manifold in $\Reals^{n}$ is
\begin{align*}
M = \Reals^{k}
  &= \{(y^{1}, \dots, y^{n}) \text{ in } \Reals^{n} : y^{k+1} = \dots = y^{n} = 0\} \\
  &= \{(y^{1}, \dots, y^{k}, 0, \dots, 0) : \text{$y^{i}$ real for each $i = 1, \dots, k$}\}.
\end{align*}
If $y \in M$ and $U$ is an arbitrary open neighborhood of $y$, the identity map from $U$ to $V = U$ satisfies the definition.
To unravel the definition in general, it's probably easier to look at it backward: Suppose you have a diffeomorphism $\phi$ defined in some open ball $V$ centered at the origin $0$ in $\Reals^{n}$. The set $M = \phi(V \cap \Reals^{k})$ is a $k$-dimensional submanifold of $\Reals^{n}$. Why? Because if we put $U = \phi(V)$ and $f = \phi^{-1}$, then $U$ is an open neighborhood of each point of $M$, and $f:U \to V$ is a diffeomorphism such that
$$
f(U \cap M)
  = V \cap \Reals^{k}
  = \{y \text{ in } V : y^{k+1} = \dots = y^{n} = 0\}.
$$
(Such a diffeomorphism $\phi$ is called a (local) parametrization of $M$; a diffeomorphism $f$ as in the definition is a (local) coordinate system on $M$.)
So why is the definition "stated backward"? In practice, you usually have a set $M$ in mind, not a parametrization whose image is of interest. If you start with a set $M$, the natural question to ask is, "If $x$ is an arbitrary point of $M$, does there exist a parametrization of some open neighborhood of $x$ in $M$?" When you make this notion precise, you've got your book's definition.
What this means is, if you have some "favorite" diffeomorphisms, you can write down lots of manifolds. You don't have any favorites? Allow me to suggest some. :)
Let $V \subset \Reals^{k}$ be a non-empty open set, and let $g:V \to \Reals^{n-k}$ be a smooth mapping. The graph of $g$,
$$
\Gamma_{g} = \{(x, y) \text{ in } V \times \Reals^{n-k} : y = g(x)\}
$$
is a smooth $k$-dimensional submanifold of $\Reals^{n}$. To see why, write $g(x^{1}, \dots, x^{k}) = (y^{1}, \dots, y^{n-k})$. The mapping $\phi:V \times \Reals^{n-k} \to V \times \Reals^{n-k}$ defined by
$$
\phi(x^{1}, \dots, x^{n})
  = (x^{1}, \dots, x^{k}, x^{k+1} + y^{1}, \dots, x^{n} + y^{n-k}), \qquad x \in V \times \Reals^{n-k}
$$
is a diffeomorphism parametrizing the graph. (If $k = 1$ and $n = 2$, we have $\phi(x^{1}, x^{2}) = \bigl(x^{1}, x^{2} + g(x^{1})\bigr)$. Be sure you understand the geometric effect of this mapping.) Points to check: $\phi$ is smooth, bijective, and has smooth inverse; the image of $V \cap \Reals^{k}$ is $\Gamma_{g}$.
One more item should be mentioned, since your definition of "manifold" is not the "intrinsic" definition used in more advanced courses.
Theorem: Let $W$ be a non-empty open subset of $\Reals^{k}$. If $g:W \to \Reals^{n}$ is smooth, proper, bijective to its image, and if the differential $Dg$ has maximal rank $k$ at each point of $W$, then the image $M = g(W)$ is a $k$-dimensional submanifold of $\Reals^{n}$.
(A mapping is proper if the preimage of every compact set is compact. This condition prevents phenomena such as wrapping an open interval to make a figure-8, or having a curve wind around on itself like a "line of irrational slope" on a torus.)
The proof entails showing that $g$ followed by some projection to a $k$-dimensional coordinate subspace is a diffeomorphism between open subsets of $\Reals^{k}$ (easy linear algebra), then extending $g$ to some open neighborhood $V$ of $W$ in $\Reals^{n}$ (clever, but natural in retrospect), and finally invoking the inverse function theorem ("the vice grips of multivariable calculus") to guarantee that the extended mapping is a diffeomorphism on some open set (possibly smaller than $V$).
A: Let $M$ be the line $x=0$, and for any point of $M$, let $U=\Bbb{R}^2$, let $f$ be the linear transformation $$f(x,y)=\pmatrix{0 & 1 \\ 1 & 0}\pmatrix{x \\ y}.$$
Then since $f$ is linear it is a diffeomorphism, and $$f(U\cap M)=f(M)=f(\{(0,y):y\in\Bbb{R}\})=\left\{\pmatrix{0 & 1\\ 1 & 0}\pmatrix{0\\y} : y\in\Bbb{R}\right\} = \{(y,0):y\in\Bbb{R}\}$$
Honestly I'm not sure this is particularly illuminating, although changing the argument slightly, this shows any line is a manifold. Essentially what the definition is saying is that for every point $x\in M$, there is some neighborhood $U$ of $x$ such that the part of the manifold near $x$ looks like some open set in a flat $k$-dimensional space $V$ (since the last $n-k$ coordinates go to 0). Where looks like is a sort of nontechnical term that here means via a diffeomorphism $f$.
