What am I doing wrong while proving time dilation using Minkowski space-time diagram? 
Consider two events A and C. They have same value of x' and time interval between them is $\Delta \tau$. This is the proper time interval between the events. The time interval between the events in the ct-x frame is $\Delta t$.
$ \Delta \tau = t^\prime_c -t^\prime_a \\
\Delta t = t_c -t_b$
Using the idea of invariant separation, I can write
$ c^2 (t_b - t_c)^2 - (x_b -x_c)^2 = c^2 (t^\prime_b - t^\prime_c)^2 - (x^\prime_b -x^\prime_c)^2 \\
c^2 (t_b - t_c)^2 - (x_b -x_c)^2 = c^2 (t^\prime_a - t^\prime_c)^2 - (x^\prime_b -x^\prime_c)^2 \\
c^2 (\Delta t)^2 = c^2 (\Delta \tau)^2 - (\Delta x^\prime)^2 \\
$
Then, If I use Lorentz transformation for $(\Delta x^\prime)^2$, we get
$ \Delta t = \frac{\Delta \tau}{\gamma}$
What am I doing wrong here? Why am I getting time contraction instead of time dilation?
 A: So, just to clarify your approach:
You take events A and C occurring at the same location $x'_a = x'_c$ but different times $t'_a$ and $t'_c$ in the primed frame. You also take event B occurring at the same time as A in the primed frame, $t'_b = t'_a$, but at the same location as C in the unprimed frame, $x_b = x_c$. For events B and C you then apply the invariance of the space-time interval
$$ 
c^2 (t_b - t_c)^2 - (x_b -x_c)^2 = c^2 (t^\prime_b - t^\prime_c)^2 - (x^\prime_b -x^\prime_c)^2
$$
and in view of $x_b = x_c$, $t'_c = t'_a$, obtain
$$
c^2 (t_b - t_c)^2 = c^2 (t^\prime_a - t^\prime_c)^2 - (x^\prime_b -x^\prime_c)^2\\
c^2 (\Delta t)^2 = c^2 (\Delta \tau)^2 - (\Delta x^\prime)^2 \\
$$
Lastly, the Lorentz transform of $\Delta x' =  x^\prime_b -x^\prime_c$ gives $\Delta x' = \gamma(x_b - v t_b) - \gamma(x_c - v t_c) \equiv \gamma v \Delta t$, and you conclude, correctly, that $c\Delta t = c\Delta \tau/\gamma$. 
Your "problem" is that regardless of the substitution $t'_c = t'_a$ your final relation still gives $(t_b - t_c) = \gamma(t'_b - t'_c)$. Let's tally up what we have: 
If we use events B and C in the unprimed frame, but A and C in the primed frame, we find that 

"The time interval between events B and C occurring at the same location in the unprimed frame appears time dilated wrt the time interval between event C and an event A occurring in the primed frame at the same location as C but at the same time as B". 

The last piece of information, "occurring in the primed frame at the same time as B", is the crucial one: we can replace event A with any other event, at any location, as long as it "occurs in the primed frame at the same time as B".
Otherwise, if we dispense with event A and simply refer to events B and C only, we just find that 

"The time interval $(t_b - t_c)$ between two events B and C occurring at the same location in the unprimed frame appears time dilated in the primed frame". 

Typical time dilation. Check!        
A: When comparing times in the context of time dilation, you are comparing the time interval between two events.  The two events must be the same two events in each reference frame.
So lets focus on the time interval between events $A$ and $C$.  By invariance of the spacetime interval:
$$-c^2 \Delta t^2 + \Delta x^2 = -c^2 \Delta {t^\prime}^2 + \Delta {x^\prime}^2. $$
In the primed reference frame $A$ and $C$ are colocated, so $\Delta x^\prime=0$, and we identify $\Delta t^\prime=\Delta\tau$ as the proper time.
$$-c^2 \Delta t^2 + \Delta x^2 = -c^2 \Delta\tau^2$$
A little algebra gets us:
$$\Delta t^2 \left[ 1 - \left(\frac{\Delta x}{c \Delta t}\right)^2 \right] = \Delta\tau^2 \implies \Delta t = \gamma\Delta\tau.$$
