What's wrong with this equal probability solution for Monty Hall Problem? I'm confused about why we should change door in the Monty Hall Problem, when thinking from a different perspective  gives me equal probability. 

Think about this first: if we have two doors, and one car behind one of them, then we have a 50/50 chance of choosing the right door. 

Back to Monty Hall: after we pick a door, one door is opened and shows a goat, and the other door remains closed. Let's call the door we picked A and the  other closed door B. Now since 1 door has already been opened, our knowledge has changed such that the car can only be behind A or B. Therefore, the problem is equivalent to: given two closed doors (A and B) and one car, which door should be chosen (we know it's a 50/50 thing)?
Then, not switching door = choosing A, and switching door = choosing B. Therefore, it seems that switching should be equally likely, instead of more likely. 

Another way to think: no matter which door we choose from the three, we know BEFOREHAND that we can definitely open a door with a goat in the remaining two. Therefore, showing an open door with a goat reveals nothing new about which door has the car. 

What's wrong with this thinking process? (Note that I know the argument why switching gives advantage, and I know experiments have been done to prove that. My question is why the above thinking, which seems legit, is actually wrong.) Thanks.
Your error is a common one: You have confused the event that the host reveals door $C$ with the event that there is a goat behind door $C$. This is distinction enough to see: if the car is, in fact, behind door $A$, then the host will only reveal door $C$ with a probability of $50\%$, whereas there is definitely a goat behind door $C$.

In particular, let $C_{\text{goat}}$ be the event that there's a goat behind door $C$, and $C_{\text{revealed}}$ be the event that door $C$ is revealed, and $A_{\text{car}}$ be the event that there is a car behind door $A$. Then, one can see that $A_{\text{car}}$ and $C_{\text{revealed}}$ are independent of each other, since the host reveals door $B$ or $C$ randomly with $50\%$ probability if there's a car behind door $A$, and if there's not a car behind door $A$, the host is forced to pick the door which hasn't got a car behind it - which is $B$ or $C$ with probability $50\%$. Thus, one concludes that
$$P(A_{\text{car}}|C_{\text{revealed}})=P(A_{\text{car}})=\frac{1}3.$$
The calculation you've done (annotated with a $\neq$ sign where things go wrong) is:
$$P(A_{\text{car}}|C_{\text{revealed}})\neq P(A_{\text{car}}|C_{\text{goat}})=\frac{1}2$$
Assume A event that selected door have prize
$P(A)$ = $\frac{1}{3}$
And B event that selected door don't have prize
$P(B)$ = $\frac{2}{3}$
Assume $S$ to be event that he switch doors and get right prize door
And $S'$ to be event that he didn't switch doors
So $P(S)$ = $P(A)$*P($\frac{S}{A}$) + $P(B)$*P($\frac{S}{B}$)
= $\frac{1}{3}$$*$$0$ + $\frac{2}{3}$$*$$1$
= $\frac{2}{3}$
$P(S')$ = $\frac{1}{3}$
Clearly we can see that the probability of winning the prize after switching is more, so switching is advisable.