How to decide which is the proper time? I want to deduce Length Contraction using Time Dilation(which has already been deduced), but I encountered a problem that I feel tough.
I first assume two observers, $A$ and $B$ are in two frames with relative speed $v$ between them, and assume just next to B, there is a stick which B measures to be of length $l$.
Now I let $B$ record the time needed for $A$ to traverse the stick, thus $t_B={l\over v}$, note that $t_B$ is the time measured in $B$'s frame using $B$'s clock.
Then I go to $A$'s frame, and I assert that $A$ will record a time $t'_A={l'\over v}$ for $A$ to traverse the stick.
Now the only thing left is to relate $t'_A$, the time measured in $A$'s frame using $A$'s clock, and $t_B$.
So my questions are:

1). How is it justified that the relative speed between the two frame is the same as measured by each observer in each frame?
2). How to use time dilation formula to relate $t'_A$  with $t_B$?

Please help me to clarify my misconception: Shall I let $B$ look at $A$'s clock in $B$'s frame, or shall I let $A$ to look at $B$'s clock in $A$'s frame?
 A: The answer to 1) is that when you say there are two observers in a relative motion $v$, Then by definition it follows that each observer claims he's at rest and it's the other observer who's moving with speed $|v|$. 
Or to put it another way, consider the relativistic velocity combination formula:
$w=\dfrac{u-v}{1-\dfrac{uv}{c^2}}$ 
$u$ is the velocity of an object as measured by a certain reference frame $S$, and $v$ is the relative velocity between $S$ and another frame $S'$. Therefore to calculate the speed of that object in the $S'$ frame which we denote as $w$, you plug in this formula.
You say that observers $A$ and $B$ are in relative motion $v$. So let's say that observer $A$ measure $B$ to be moving with $v$ in the positive x-direction.The object which we are interested in calculating its speed is observer $A$. In his frame, observer $A$ will claim he's at rest so that we have $u=0$. Plugging in the above formula:
$v_\text{of A as measured by B} =\dfrac{0-v}{1-\dfrac{0*v}{c^2}}=-v$
So this is the justification. If two observers are in relative motion, they measure the same speed for the other observer.
This conclusion in fact has nothing to do with special relativity and it works the same if we used the ordinary velocity combination formula: $w=u-v$. This differs from the relativistic equation by a factor of $1-\dfrac{uv}{c^2}$, but it gives the same result.
As to you second question, I'll give you a hint:
$t_o=\dfrac {t_\text{moving}} {\sqrt{1-(\dfrac{v}{c})^2}}$
This is the equation of time dilation that relates two clocks, $t_o$ and $t_\text{moving}$, which is at rest and which is moving with velocity $v$, respectively.
Try to use this formula in your second equation $t'_A={l'\over v}$ and relate $t'_A$ to $t_B$ to derive the equation for length contraction. 
A: Another take on Omar Nagib's Answer is that one justifies the use of velocity $-v$ for A relative to B whenever we use $v$ for B relative to A by what is sometimes (rather vaguely and unhelpfully) the reciprocity relationship.
As Omar notes this relationship is not particular to special relativity but also applies to the Galilean velocity addition (infinite $c$) relativity. And the fundamental reason is this:
Spatial Isotropy
Galileo's principle (of indetectability of uniform motion of a frame from within a frame) implies the transformations between inertial frames together with transformation composition form a group. Copernicus's principle that spacetime is homogeneous implies that the transformations act linearly on affine co-ordinates. So we have a matrix group action whereby 
$$X=\left(\begin{array}{c}x\\t\end{array}\right)\mapsto \exp(\eta\,K)\,\left(\begin{array}{c}x\\t\end{array}\right)\tag{1}$$
where $K$ and $\exp(\eta\,K)$ are $2\times 2$ matrices and $\eta$ encodes the relative motion's "swiftness" (think of it as a generalized velocity - an invertible function of the time over distance velocity).
Now, by spatial isotropy, we can now choose the $-x$ direction to be our positive spatial direction. There's nothing fundamentally different about this direction compared with the $+x$ direction, so if we reason above, we must get the same $K$ matrix as above, and a possibly different $\eta$ parameter, say $\eta^\prime$.
Thus, when we make the transformation $x\mapsto-x$, we must have:
$$\exp(\eta^\prime\,K)\left(\begin{array}{c}-x\\t\end{array}\right)=\exp(\eta\,K)\left(\begin{array}{c}+x\\t\end{array}\right)\tag{2}$$
for all $x$ and $t$. A little bit of algebra then shows that $\eta^\prime= -\eta$ and, on "calibrating" the $\exp(\eta\,K)$ in terms of everyday velocities, we find $v^\prime = -v$.
You might like to read some more details on this in my article "Of Groups, Galileo and what's so special about the speed of light": I've put a preprint on my website and hope it will be published in the European J. Physics sometime soon. "Reciprocity" is also discussed in:
Stefano Liberati, Sebastiano Sonego, Matt Visser, "Faster-than-c signals, special relativity, and causality", Ann. Phys. 298 (2002) pp167-185
A: Simpler setup. Have two people with a meter stick between them, all mutually at rest.
The are in the same frame so they synchronize their watches. Then at the same watch time they send a signal to each other and hold a mirror and bounce the other one's signal back and then when their own light comes back they look at their watch.
If their watch says it took $\Delta t$ seconds, then they know they must be a distance $L$ apart where $2L=c\Delta t.$
So far, all in one frame. Now consider a new frame, according to time dilation it thinks those clocks were moving so ticking slower so it thinks the time between sending and receiving was longer, the new frame thinks it took time $\Delta T.$
In the new frame light takes off towards an end of the rod that is a distance $l$ away moving at a speed $v$ then bounces back and travels towards the original end of the rod a distance $l$ rushing towards it at speed $v.$
Let's figure out those times. When it is rushing away from you with a head start of $l$ then if it takes time $T_1$ to meet then $T_1c=l+vT_1$ or $T_1=l/(c-v).$ When you were rushing towards each other starting $l$ away then if it takes time $T_2$ to meet then $l=cT_2+vT_2$ or $T_2=l/(c+v).$
So the time it takes to go there and back is $\Delta T=T_1+T_2$ which equals:
$\begin{eqnarray}\Delta T &= &\frac{l}{c-v}+\frac{l}{c+v} \\
&= &\frac{l(c+v)}{c^2-v^2}+\frac{l(c-v)}{c^2-v^2} \\
&= &\frac{2lc}{c^2-v^2}.
\end{eqnarray}$
Now $\Delta T/\Delta t$ is known and it equals $\frac{2lc/(c^2-v^2)}{2L/c},$ which equals $\frac{lc^2}{L(c^2-v^2)}$ which means $l/L=(\Delta T/\Delta t)(c^2-v^2)/c^2.$
Now from time dilation we know that $\Delta T/\Delta t=\sqrt{c^2/(c^2-v^2)}$ so we get $l/L=\sqrt{(c^2-v^2)/c^2}$ or that the rod that was length $L$ to the frame it was at rest in is of length $l=\frac{L}{\sqrt{1/(1-(v^2/c^2))}}.$
