How do I make a student understand contradiction?

We were trying to prove that if $3p^2=q^2$ for nonnegative integers $p$ and $q$, then $3$ divides both $p$ and $q$. I finished writing the solution (using Euclid's lemma) when a student asked me

"How can you assume $3p^2=q^2$ when that implies $\sqrt 3$ is rational which we know is false?"

I told him that question at hand is used as a lemma in proving that $\sqrt 3$ is irrational. But he gave another argument using fundamental theorem of arithmetic to independently prove that $\sqrt 3$ is irrational. So I could not convince him that one can give a chain of arguments starting from a hypothesis without actually knowing the truth value of the hypothesis itself. Later I came back and tried to think more about this a bit fundamentally to understand what it means to prove the implication $"A \to B"$ without worrying about truth of $A$ and how different is it from proving $B$ when we know $A$ to be true (or false, as in this case, using some other method). But I am also confused. Another student made this remark, "you are giving me one false statement, and asking me to prove another false statement, I don't understand". Can someone please resolve this?

Proof by contradiction is simple. It is not so much that if a false statement is true, then we arrive at a contradiction. Rather, a better way to think of it is, "if a given statement is true--a statement whose truth or falsity is not yet established--then we arrive at a contradiction; thus the original statement cannot be true."

Take your proof of the irrationality of $\sqrt{3}$ as an example. At the outset, we have not yet established whether or not $\sqrt{3}$ is irrational (or rational). And we don't know until the proof is complete. But you certainly can reason about it by first supposing that if $\sqrt{3}$ were rational, then there exist two positive integers $p, q$ such that $q$ does not divide $p$, for which $p/q = \sqrt{3}$. That follows from the supposition of rationality. Then by continuing the logical chain, you arrive at a contradiction. So if all the consequential steps from the original supposition are valid, but the conclusion is inconsistent, then the original supposition could not be true. Note that in this chain of reasoning, at no point do we actually claim that the original supposition is true, because as we have taken care to mention, we do not know if it is or not until the proof is complete. We are merely saying that if it were the case, we would arrive at a contradiction.

Here is a less trivial example. Some conjectures remain open in the sense that we do not know what the answer is. Consider the Collatz recursion $$x_{n+1} = \begin{cases} 3 x_n + 1, & x_n \text{ is odd} \\ x_n / 2, & x_n \text{ is even}. \end{cases}$$ The conjecture is that for every positive integer $m$, the sequence $\{x_n\}_{n=1}^\infty$ with $x_0 = m$ contains $1$ as an element in the sequence.

If we were to attempt to prove this conjecture is TRUE by contradiction, the supposition we would presumably start with is that there exists a positive integer $m^*$ such that if $x_0 = m^*$, the sequence $\{x_n\}_{n=1}^\infty$ never contains $1$, and then by using some as-yet-unknown logical deduction from this supposition, if we can arrive at a contradiction, we then have shown that the conjecture is true.

But note what I said above: this is an open question! Nobody knows if it is true or false. So you cannot claim that my logic is invalid because I have a priori assumed something that I already know not to be the case!

Now, if I were to desire to disprove the conjecture, it would suffice to find just such an instance of $m^*$ as I have described above. But an extensive computer search has turned up no such counterexample. This leads many to believe the conjecture is true, but no constructive method to substantiate that belief. Mathematics--and number theory in particular--contains many examples of claims that appear true but have their smallest known counterexamples for very large numbers.

So, there is my "informal" explanation of proof by contradiction. The essence is that we don't presume to know the truth or falsity of a given claim or statement at the outset of the proof, so we cannot be accused of assuming that which is false. The falsity is established once the contradiction is shown.

Step 0. Use truth tables to convince him that $A \rightarrow \bot$ is equivalent to $\neg A$. Deduce that to prove $\neg P$, we can assume $P$ and attempt to derive $\bot$, since this allows us to infer $P \rightarrow \bot$, which is the same as $\neg P$.

Step 1. Use truth tables to convince him that $A \rightarrow B$ is equivalent to $\neg(A \wedge \neg B).$ Conclude that to prove $A \rightarrow B$, we may assume $A \wedge \neg B$ and attempt to deduce $\bot$.

Edit. My original answer (above) focuses on the logic of proof by contradiction, but there is also a pedagogical issue here that deserves to be addressed. The following material is taken from the user Keen and from Steve Jessop's excellent answer.

Unfortunately, assume is one of those words where the mathematical definition differs from its popular English definition. In English, assume S means believe S. In mathematics, it means only imagine if S were true. So when we say, "assume $3p^2=q^2$", we are not asserting that there exists an ordered pair $(p,q)$ with this property. Rather, we are considering the hypothetical properties that such a pair would have, if it did exist.