Prove Logic Using Proof of Contradiction Prove the following using proof of contradiction:
 a → c 
 b → d 
 (c ∨ d) → ¬e 
 (e ∨ f) → (a ∨ b) 
 ∴ ¬e

What I have done so far is:
 1. a → c 
 2. b → d 
 3. (c ∨ d) → ¬e 
 4. (e ∨ f) → (a ∨ b) 
 5. ¬(¬e)  Assume for contradiction
 6. ¬¬e  De Morgan (5)
 7. e Double Negation (6)
 8. ¬a ∨ c  Implication Equivalence (1)
 9. ¬b ∨ d  Implication Equivalence (2)

What do I do next? Please help
 A: In the proof the OP offered are these lines:


  
*¬(¬e)  Assume for contradiction  
  
*¬¬e  De Morgan (5)
  

De Morgan's laws would not justify going from line 5 to 6. Rather one can ignore the parentheses in line 5 and write it as $¬¬e$.
The following is a proof using a Fitch-style proof checker. It does not have material implication that was used by the OP in lines 8 and 9 but approaches the proof using modus tollens (MT), conjunction elimination (∧E), conjunction introduction (∧I), De Morgan's laws (DeM), double negative elimination (DNE), disjunction introduction (∨I), negation elimination (¬E) and indirect proof (IP)


Kevin Klement's JavaScript/PHP Fitch-style natural deduction proof editor and checker http://proofs.openlogicproject.org/
P. D. Magnus, Tim Button with additions by J. Robert Loftis remixed and revised by Aaron Thomas-Bolduc, Richard Zach, forallx Calgary Remix: An Introduction to Formal Logic, Fall 2019. http://forallx.openlogicproject.org/forallxyyc.pdf
A: Well, under the assumption that $e$, we know that  $$(e\vee f)\to(a\vee b) = (\lnot e\wedge \lnot f)\vee a\vee b$$Since $e$, $$(e\vee f)\to(a\vee b) =a\vee b = c\vee d$$But, $$c\vee d\to \lnot e$$ A contradiction. So, $\lnot e$.
A: Tip: Just assume $\rm e$, so should you manage to derive a contradiction, there you may use negation introduction to immediately arrive at the desired conclusion.
Having assumed $\rm e$, look for a premise in which it occurs.  That would be premise 4, where it is a disjunct in the antecedant of a conditional.  Introduce a disjunction and eliminate the condtional.
Having so derived $\rm a\lor b$, you need to ... complete the proof by filling in the missing details.
|  1. a → c 
|  2. b → d 
|  3. (c ∨ d) → ¬e 
|_ 4. (e ∨ f) → (a ∨ b) 
|  |_ 5. e
|  |  6. e ∨ f     disjunction introduction, 5
|  |  7. a ∨ b     conditional elimination, 6, 4
:  :  
|  | 19. ¬e        disjunction elimination, 7, ..., ...
|  | 20. #         negation elimination, 5, 19
| 23. ¬e           negation introduction, 5-20


$$\def\fitch#1#2{\quad\begin{array}{|l}#1\\\hline#2\end{array}}{\fitch{~~1.~a \to c \\~~2.~b \to d \\~~3.~(c \vee d) \to \neg e \\~~4.~(e \vee f) \to (a \vee b)}{\fitch{~~5.~e}{~~6.~e\vee f\hspace{10ex}\vee\mathsf I,5\\~~7.~a\vee b\hspace{10ex}\to\mathsf E,6,4\\\fitch{~~8.~a}{~~9.~c\hspace{10ex}\to\mathsf E, 8,1\\10.~c\vee d\hspace{6ex}~\vee\mathsf I, 9}\\11.~a\to c\vee d\hspace{5ex}\to\mathsf I, 8{-}10\\\fitch{12.~b}{13.~d\hspace{10ex}\to\mathsf E, 12,2\\14.~c\vee d\hspace{7ex}\vee\mathsf I, 13}\\15.~b\to c\vee d\hspace{6ex}\to\mathsf I, 12{-}14\\16.~c\vee d\hspace{11ex}\vee\mathsf E, 7,11,15\\17.~\neg e\hspace{13ex}\to\mathsf E, 16, 3\\18.~\bot\hspace{14ex}\neg\, \mathsf E,5,17}\\19.~\neg e\hspace{17ex}\neg\,\mathsf I, 5{-}18}\\\blacksquare}$$

