How to know which hypothesis is failing? We always do a lot of proofs by contradiction. We suppose $P$, then we get to a contradiction, like $Q \wedge \neg Q$, and thanks to this, we can say that $\neg P$ is true. 
The problem is that, implicitly, we are always assuming lots of statements, not only $P$ (whenever we say, e.g. "let $V$ be a subspace", we are assuming that). How can we always say that the statement which caused the absurdity was $P$ and not a hypothesis for the problem?
As an example: 

Let $W_1, W_2$ be subspaces of $V$, such that $W_1 \subset W_2$ or $W_2 \subset W_1$, and suppose that $W_1 \cup W_2$ is not a subspace of $V$ (... insert a path to absurdity here ...), then, by RAA we know that $W_1 \cup W_2$ is a subspace of $V$

How do we know here that the assumption failing is that $W_1 \cup W_2$ is not a subspace, instead of being one of the others? (that $W_1 \subset W_2$, or $W_2 \subset W_1$, or even that $W_1, W_2$ are subspaces).
 A: This is a good question; I think a slightly abstract example will help, but first let me address your specific example and hopefully give some flavor of the answer:

Think about what you're trying to do. 
In your example, what we want to prove is "If $W_1, W_2$ are subspaces where one contains the other, ..." - so our proof by contradiction is going to occur inside a "bigger" proof where we begin by assuming our hypothesis and end by concluding the thing we want to show (that $W_1\cup W_2$ is a subspace).
Our conclusion reflects this; we don't prove "$W_1\cup W_2$ is a subspace," but rather "If $W_1, W_2$ are subspaces where one contains the other, then $W_1\cup W_2$ is a subspace."

OK, now let me be a bit more abstract.
Suppose I have sentences $A$ and $B$ (for example, $A=$"$W_1, W_2$ are a vector space with one contained in the other") and I have a proof of a contradiction from "$A$ and $B$;" I'm going to ignore exactly what $A$ and $B$ are right now, or how that proof goes - just suppose it's a thing I can do. Then there are really three proofs kicking around here, each of which is correct:


*

*Proof $1$: Assume $A\wedge B$, get a contradiction; conclude "it is not the case that both $A$ and $B$ are true."

*Proof $2$: Assume $A$. Assuming $A$, now assume $B$ and get a contradiction; conclude, assuming $A$, get not $B$. Now "step outside" of this "inner proof" and conclude "if $A$ is true, then $B$ is false.""

*Proof $3$: Assume $B$. Assuming $B$, now assume $A$ and get a contradiction; conclude, assuming $B$, "not $B$." Now "step outside" of this "inner proof" and conclude "if $B$ is true, then $A$ is false."
(It's a good exercise in propositional logic to check that $\neg(A\wedge B)$, $A\implies\neg B$, and $B\implies\neg A$ are all equivalent.) Note that the "unexamined" hypotheses in $(2)$ and $(3)$ - namely, $A$ and $B$ respectively - don't vanish; they stick around as hypotheses in the conclusion. More generally, note that these three proofs prove different things; your "which hypothesis is wrong?" question boils down to "which conclusion of $(2)$ or $(3)$ do we actually want?" (noting that both are correct).
What's going on is that proofs have structure, and one thing we can do in a proof is introduce a subproof. This is how a proof by contradiction works, in general; we want to prove a statement of the form $A\implies C$, not just $C$ alone, so if we can get a contradiction from $A\wedge B$ what we really want is the form "$A\implies$ not $B$," so that we can use "not $B$" going forward in our attempt to prove $C$.
A: Good question. Yes, when some assumption $P$ leads to a contradiction it is typically within the context of having made a bunch of other assumptions $T$ (I use $T$, since typically we are working within some theory), so when we get a contradiction, then, at least purely logically, it seems the best we can say is that $T \cup P$ is an inconsistent set of statements.  Indeed, when we point to the proof by contradiction and say 'this is a proof that $P$ is false', what we should really be saying is that 'If $T$ is true theory, then $P$ is false'.
But like I said, that is from a purely logical point of view. In practice, we certainly 'prefer' the truth of certain statements over others, sometimes for the simple reason that we have evidence that certain statements are true in the real world. 
So, for example, take the classic argument that the Earth is round, given the observation that ships that sail off to the horizon gradually disappear, from bottom to top. Now, that certainly seems to disprove the Flat Earth Hypothesis, as the Flat Earth Hypothesis would predict that we should keep seeing the whole ship, i.e. the Flat Earth Hypothesis leads to a prediction that contradicts our observation.  However, the Flat Earth Hypothesis proponents could say: 
"Well, please know that the prediction that you should see the whole ship is made within the context of all kinds of other hypotheses, including the one that says that light goes in a straight line. So, rather than rejecting the Flat Earth Hypothesis , maybe we should reject the hypothesis that light goes in a straight line: maybe light gets 'repelled' by the Earth and, as such, we would indeed see the ship gradually disappearing given the Flat Earth Hypothesis!"
OK, yes, we could do that ... but of course we have far more reasons (through other observations and arguments) to believe that light does go in a straight line than that the Earth is flat.
And the same really goes for math. Yes, working within the Theory of Peano Arithmetic, say, we could say: "Well, the hypothesis that there is a greatest prime number leads to a contradiction, yes, but rather than concluding that there is no greatest prime number, maybe we should reject one of the Peano Axioms".   And sure, sometimes interesting new theories come to be exactly by rejecting certain long-standing hypothesis (non-Euclidian geometry is a good example), but such cases are pretty rare: there is a fairly stable set of axioms that we typically hold in mathematics since they have great applicability to the real world, and so we'd much rather say: "Given that $P$ leads to a contradiction, $P$ is false (and so implicitly we can hang on to our core set)" than "$P$ leads to a contradiction, so let's replace one of our core mathematical assumptions with $P$"
A: Noah Schweber's answer is good, but I want to actually present the concept in the a sequent-like notation, as I find this approach one of the best ways of understanding what's going on in a proof. 
We rarely want to prove tautologies, usually we want to prove a statement given some other statements. We often write this as $\varphi,\psi\vdash \chi$ for "$\chi$ is provable given (proofs for) $\varphi$ and $\psi$". The list of assumptions, $\varphi,\psi$ here, is often represented by $\Gamma$ and called the context. For your example, the context, i.e. $\Gamma$ contains the statements "$W_1$ is a subspace of $V$", "$W_2$ is a subspace of $V$", and "$W_1\subseteq W_2 \lor W_2\subseteq W_1$".
The rule for what you're calling RAA (which is proving $P$ by finding a contradiction from $\neg P$, which is different from proving $\neg P$ by finding a contradiction from $P$) is the following: $$\cfrac{\Gamma,\neg\varphi\vdash \bot}{\Gamma\vdash\varphi}RAA$$ This says that if we can prove a contradiction ($\bot$) given a context containing $\neg\varphi$, then via RAA we can prove $\varphi$ from the context without $\neg\varphi$. For your example, $\Gamma$ is as before, and $\varphi$ is the statement "$W_1\cup W_2$ is a subspace of $V$". Your elided reduction to a contradiction would go above the use of RAA, i.e. $$\cfrac{\cfrac{\vdots}{\Gamma,\neg\varphi\vdash \bot}}{\Gamma\vdash\varphi}RAA$$
How does RAA know to apply to $\varphi$? Because we tell it. In this case by using a positional approach i.e. RAA applies only to the rightmost assumption in the context. (We have so-called structural rules that let us rearrange the order of assumptions in the context. These are typically elided in semi-formal proofs.) Why can't we pick on say "$W_1\subseteq W_2 \lor W_2\subseteq W_1$", call it $\psi$, instead of $\varphi$? We can. We'd use a different rule from RAA, namely $\neg$-introduction which looks like: $$\cfrac{\Gamma,\varphi\vdash\bot}{\Gamma\vdash\neg\varphi}\neg I$$
With this we can prove $\Gamma'\vdash\neg\psi$ where $\Gamma'$ is the context containing $\neg\varphi$ but omitting $\psi$. This, however, would be the statement, "given W_1, W_2 are subspaces of V and $W_1\cup W_2$ is not, then we can prove that neither $W_1\subseteq W_2$ nor $W_2\subseteq W_1$," which is not what we're trying to prove.
As the name "context" suggests, the context holds the background assumptions/definitions about which we're reasoning. Typically there is no hard separation between background assumptions and temporary "working" assumptions in natural deduction/sequent calculus presentations. You could imagine using some notation to emphasize this, e.g. $\Gamma;\varphi\vdash\psi$. In fact, focused systems do make a harder distinction between "background" assumptions and "working" assumptions and require explicitly copying background assumptions into the "working space" before they can be used, i.e. they have a rule that looks like $\cfrac{\Gamma,\varphi;\Delta,\varphi\vdash\psi}{\Gamma,\varphi;\Delta\vdash\psi}$ where $\Delta$ is a list of "working" assumptions.
