Proof that $\sqrt{2}$ is irrational is not convincing. Please help. I understand the irrationality of $\sqrt{2}$ in the following way:
To prove: $\sqrt{2}$ is irrational
Proof: Assume $\sqrt{2}$ is rational.
i.e. $\sqrt{2}=\dfrac{a}{b}$
Assume $a$ and $b$ are co-prime
...... (the usual steps)
Hence $a$ and $b$ cannot be co-prime.
This contradicts our second assumption.
So first assumption is wrong.
So $\sqrt{2}$ is irrational.

MY CONFUSION:
We are making two different assumptions. This is not the way proof by contradiction works. If the second assumption gets contradicted, for what reason will the first assumption be false?

 A: The first assumption implies the second. If a number is rational, then we can write it as $\frac ab$ with $a,b$ coprime. Thus if we cannot write a number as $\frac ab$ with $a,b$ coprime, then it is not rational.
A: This works since if $q$ is assumed to be rational, then it's always true that it can be expressed as a fraction of coprime integers. Thus if you disprove this, it means that $q$ is not rational.
A: The second assumption can be avoided if it bothers you. You can say:
Suppose $\sqrt 2=\frac{a}{b}$ ($a,b$ not necessarily co-prime).
Let $c=\frac{a}{(a,b)}$ and $d=\frac{b}{(a,b)}$, where $(a,b)$ denotes the highest common factor of $a$ and $b$.
Then $\sqrt 2=\frac{c}{d}$, and $c$ and $d$ are co-prime.
Now proceed with the proof as above, with $c,d$ in place of $a,b$.
A: Well, you can always assume that in the quotient of integers, $q=a/b$, numerator and denominator are coprime.
The proof makes use of the main theorem of arithmetic that each positive integer can be uniquely written as a product of prime powers. This will lead of a contradiction:
If $\sqrt 2 = \frac{a}{b}$ then $2 =\frac{a^2}{b^2}$, i.e., $2b^2 = a^2$. Now think about $2$ as a prime factor on the left-hand and the right-hand side and you will have the contradiction.
