Prove :(P → Q) ∨ (Q → P) using natural deduction Allowed inference rules: ∨-I, ∨-E, ∧-I, ∧-E, →-I, →-E, ¬-I, ¬-E
I tried to prove a contradiction by assuming $¬ ((P → Q) ∨ (Q → P))$ but got stuck, or am I doing it in the wrong way?
Edit:
My proof attempt
$1.\qquad¬ ((P → Q) ∨ (Q → P))\qquad \qquad\qquad\qquad\qquad\qquad\qquad \qquad\qquad Assum$
$2.\qquad P → Q\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\quad\  Assum$
$3.\qquad(P → Q) ∨ (Q → P)\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\quad\; ∨-I(2)$
$4.\qquad(¬ ((P → Q) ∨ (Q → P))) ∧ ((P → Q) ∨ (Q → P))\qquad\qquad\qquad\; ∧-I(1,3)$
$5.\qquad  ¬ (P → Q)\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\quad\;\;\; ¬-I(2,4)$
$6.\qquad  Q → P\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad \qquad\qquad\qquad\quad\; Assum$
$7.\qquad  (P → Q) ∨ (Q → P)\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\quad\;\; ∨-I(6)$
$8.\qquad  (¬ ((P → Q) ∨ (Q → P))) ∧ ((P → Q) ∨ (Q → P))\qquad\qquad\qquad\;\; ∧-I(1,7)$
$9.\qquad  ¬ (Q → P)\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\; ¬-I(6,8)$
$10.\qquad ¬ (P → Q) ∧ ¬ (Q → P)\qquad\qquad\qquad\qquad\qquad\qquad\;\qquad\qquad\quad\;\;\, ∧-I(5,9)$
Stuck at this line, can't spot any contradiction to line 1

Your start is good:
$1.\qquad¬ ((P → Q) ∨ (Q → P))\qquad \qquad\qquad\qquad\qquad\qquad\qquad \qquad\qquad Assum$
$2.\qquad \qquad P → Q\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\quad\  Assum$
This second assumption is good too, since you basically want to do a DeMorgan on line 1, and to do those in Natural Deduction, you do a Proof by COntradiction on each of the disjuncts .. so yes, assume the disjunct, but then proceed with getting the full disjunct:
$3.\qquad\qquad (P → Q) ∨ (Q → P) \qquad \qquad\qquad\qquad\qquad\qquad\qquad \qquad\qquad  \lor Intro \: (2)$
and now you can get a contradiction with line 1:
$4.\qquad \qquad\bot \qquad \qquad\qquad\qquad\qquad\qquad\qquad \qquad\qquad \qquad \qquad\qquad\qquad \bot \: Intro \: (1,3)$
So you can close the subproof and get:
$5.\qquad \lnot(P → Q) \qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad  \lnot Intro \: (2-4)$
OK, same pattern with the other disjunct, i.e. repeat those same 4 lines:
$6.\qquad \qquad Q → P\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\quad\  Assum$
$7.\qquad \qquad (P → Q) ∨ (Q → P) \qquad \qquad\qquad\qquad\qquad\qquad\qquad \qquad\qquad \lor Intro \: (6)$
$8.\qquad\qquad \bot \qquad \qquad\qquad\qquad\qquad\qquad\qquad \qquad\qquad \qquad \qquad\qquad\qquad \bot \: Intro \: (1,7)$
$9.\qquad \lnot(Q → P) \qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad  \lnot Intro \: (6-8)$
OK, we know that $\lnot (P \rightarrow Q) \equiv P \land \lnot Q$, so we should be able to get $P$ from 5, and $\lnot P$ from 9, giving us the desired contradiction. How do we prove that? It's not too hard: two more proofs by contradiction! That is, to get $P$, assume the opposite:
$10.\qquad \qquad\lnot P \qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\quad\  Assum$
And since we want this to contradict with $\lnot (P \rightarrow Q)$, try and prove $P \rightarrow Q$, which we do with a conditional proof:
$11.\qquad\qquad\qquad P \qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\  Assum$
$12.\qquad \qquad\qquad\bot \qquad \qquad\qquad\qquad\qquad\qquad\qquad \qquad\qquad \qquad \qquad\qquad \bot \: Intro \: (10,11)$
$13.\qquad \qquad\qquad\ Q \qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad  \bot \: Elim \: (12)$
$14.\qquad\qquad P → Q\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\quad\  \rightarrow \: Intro (11-13)$
$15.\qquad\qquad \bot \qquad \qquad\qquad\qquad\qquad\qquad\qquad \qquad\qquad \qquad \qquad\qquad\qquad \bot \: Intro \: (5,14)$
$16.\qquad \lnot \lnot P \qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad  \lnot \: Intro \: (10-15)$
$17.\qquad P \qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad  \lnot \: Elim \: (16)$
Almost there! Do a similar pattern to get $\lnot P$ from 9:
$18.\qquad \qquad P \qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\quad\  Assum$
$19.\qquad\qquad\qquad Q \qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\  Assum$
$20.\qquad \qquad\qquad\ P \qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad  Reit \: (18)$
$21.\qquad\qquad Q → P\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\quad\  \rightarrow \: Intro (11-13)$
$22.\qquad\qquad \bot \qquad \qquad\qquad\qquad\qquad\qquad\qquad \qquad\qquad \qquad \qquad\qquad\qquad \bot \: Intro \: (9,21)$
$23.\qquad \lnot P \qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad  \lnot \: Intro \: (18-22)$
And now we're done:
$24.\qquad \bot \qquad \qquad\qquad\qquad\qquad\qquad\qquad \qquad\qquad \qquad \qquad\qquad\qquad\qquad \bot \: Intro \: (17,23)$
$25.\lnot \lnot ((P → Q) ∨ (Q → P))\qquad \qquad\qquad\qquad\qquad\qquad\qquad \qquad\qquad\qquad \lnot Intro (1-24)$
$26. (P → Q) ∨ (Q → P)\qquad \qquad\qquad\qquad\qquad\qquad\qquad \qquad\qquad\qquad \lnot Elim (25)$
OK, so it turns out to be a somewhat lengthy proof, but that often happens with Natural Deduction, where we don;t have handy dandy equivalence rules. However, notice that between the set-ups for conditional proof, conditional proof, etc, as well as the repeating patterns for doing things like Demorgan, you end up cranking out these proofs on automatic pilot once you have done them for a while. And finally, oftentimes the instructor, book, or software will start to allow you to simply refer to those patterns as Lemma's, so you don't have to write them out every time.
