Let $P$ be the statement '$1=1$'.
Assume $\neg P$ is true (that is, assume that $1\not=1)$
Okay; let's call this Assumption (1).
and clearly $Q\land\neg Q$ is false,
so $\neg P\implies (Q\land\neg Q)$ is false,
and by logical equivalence $P$ is also false.
Yes.
I have proved that the statement $P$ is false.
No, your proof is arguing that $$\lnot P\implies(\lnot P \Longrightarrow \bot)\text{ is false}\implies P\text{ is false}.$$ You have merely proven that under Assumption (1), statement $P$ is false; in other words, you have proven this: $$1\ne1\implies (1=1\text{ is false}).\quad✅$$
This conclusion obviously isn't illuminating or useful; nonetheless, your proof is entirely valid (coherent).
Clearly, a problem has arisen, since $P$ is not false.
No problem has arisen, because you have not actually proven this:
$$1=1\text{ is false}.$$
However, can proof by contradiction be used to prove that $P$ is false?
Sure, proof by contradiction can soundly disprove a false statement.
Let $P$ be $1=2.$
Assume, for the sake of contradiction, that $\boldsymbol P$ is true.
Now, since $1<2,$ we have that $1\ne2.$
Therefore, we have the contradiction $1=2$ and $1\ne2.$
Hence, our assumption must be false: $\boldsymbol P$ must in fact be false.
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Thanks for you response! I see your point that everything I have proven is under assumption (1), which leads to an unsound argument.
Your argument/proof is actually sound, as that assumption $\color\red{1≠1}$ is not a genuine premise, since it was eventually discharged by becoming the antecedent of the proof's tautological conclusion $$\color\red{\lnot P}\implies (P\text{ is false}).\quad✅$$
I've noticed that the first step for any proof by contradiction is to assume ¬P is true. Then we prove that P is true, is that under the assumption ¬P is true as well?
No: by the end of the proof by contradiction, the assumption $¬P$ will also have been discharged. In Structure of a Proof by Contradiction, I explained that such a proof is implicitly premised on the logical entailment $$\color{green}{(\lnot P\implies \bot)}\implies P,\tag{*}$$ which is why (and how) the meat $$\color{green}{\lnot P\implies \bot}\tag{#}$$ of the proof logically implies (and allows us to finally conclude) that $P$ must in fact be true.
Side note: while sentence $(*)$ is true regardless of $P$'s truth, sentence $(\#)$ is true only if $P$ is actually true. That is to say, proof by contradiction isn't ever able to prove a false statement or disprove a true statement.
The textbook that we are using justifies proof by contradiction by stating that the statement $P$ is logically equivalent to the statement $(\neg P\implies\bot).$
To be precise: proof by contradiction doesn't actually depend on the full equivalence $$(\neg P\implies\bot)\color\red\iff P,$$ only its forward direction $(*).$
