How do I prove something without premises in a Fitch system? If asked “Prove in Fitch: From no premises, derive $A \lor (A \to B)$. Without using Taut Con?"
These are the are the Fitch rules, and this is what I have so far. 
Should I aim to use V Elim to isolate both sides and then derive with the method I'm currently trying? I'm unsure how to piece that part together.
 A: In order to:

prove something without premises

we have to take care to discharge all the "temporary" assumptions we made in the derivation.
We can prove your formula using LEM, that in turn is derivable from Double Negation.
1) $A$ --- assumed [a]
2) $A \lor \lnot A$ --- from 1) by $\lor$-intro
3) $\lnot (A \lor \lnot A)$ --- assumed [b]
4) $\bot$ --- $\bot$-intro: from 2) and 3)
5) $\lnot A$ --- by $\lnot$-intro from 1) and 5), discharging [a]
6) $A \lor \lnot A$ --- from 5) by $\lor$-intro
7) $\bot$ --- $\bot$-intro: from 3) and 6)
8) $A \lor \lnot A$ --- from 3) and 7) by DN, discharging [b]
Note: up to now we have proved $\vdash A \lor \lnot A$; this is an example of how to derive a valid formula, i.e. how to prove something without assumptions.
9) $A$ --- assumed [c] from 8) by $\lor$-elim
10) $A \lor (A \to B)$ --- from 9) by $\lor$-intro
11) $\lnot A$ --- assumed [d] from 8) by $\lor$-elim
12) $A$ --- assumed [e]
13) $\bot$ --- $\bot$-intro: from 11) and 12)
14) $B$ --- $\bot$-elim: from 13)
15) $A \to B$ --- from 12) and 14), discharging [e]
16) $A \lor (A \to B)$ --- from 15) by $\lor$-intro

17) $A \lor (A \to B)$ --- from 9)-10) and 11)-16) and 8) by $\lor$-elim, discharging [c] and [d].

A: The most direct way to do this is to start by assuming $\neg (A\: \vee \: (A\: \rightarrow \: B))$, eventually discharging it via negation introduction. The only other rules we have that allow us to discharge premises are conditional rules, but we want to end up with a disjunction, and the system itself doesn't give us a way of translating between them. 
$
\begin{array}
\\1. & \mid \: \neg (A\: \vee \: (A\: \to \: B) & \text{ass}
\\2. & \mid \: \mid \: \neg A & \text{ass}
\\3. & \mid \: \mid \: \mid \: A & \text{ass}
\\4. & \mid \: \mid \: \mid \: \mid \: \neg B & \text{ass}
\\5. & \mid \: \mid \: \mid \: \mid \: A\: \wedge \: \neg A & \text{2, 3 $\wedge$intro}
\\6. & \mid \: \mid \: \mid \: \neg \neg B & \text{4-5 $\neg$intro}
\\7. & \mid \: \mid \: \mid \: B & \text{6, $\neg$elim}
\\8. & \mid \: \mid \: A\: \to \: B & \text{3-7 $\to$intro}
\\9. & \mid \: \mid \: A\: \vee \: (A\: \to \: B) & \text{8, $\vee$intro}
\\10. & \mid \: \mid \: \neg (A\: \vee \: (A\: \rightarrow \: B)\: \wedge \: (A\: \vee \: (A\: \rightarrow \: B))\quad & \text{1, 9 $\wedge$intro}
\\11. & \mid \: \neg \neg A & \text{2-10 $\neg$intro}
\\12. & \mid \: A & \text{11, $\neg$elim}
\\13. & \mid \: A\: \vee \: (A\: \rightarrow \: B) & \text{12, $\vee$intro}
\\14. & \mid \: \neg (A\: \vee \: (A\: \rightarrow \: B)\: \wedge \: (A\: \vee \: (A\: \rightarrow \: B)) & \text{1, 13 $\vee$intro}
\\15. & \neg \neg (A\: \vee \: (A\: \rightarrow \: B)) & \text{1-14 $\neg$intro}
\\16. & A\: \vee \: (A \ \rightarrow \ B) & \text{15, $\neg$elim}
\end{array}
$
In 2-9, I show $\neg A\:\vdash \: A \vee(A\rightarrow B)$, and 12-13 show $A\:\vdash \: A \vee(A\rightarrow B)$. If you already had $A \vee \neg A$, you could use these pieces with disjunction elimination, as you suggested.
Make sure to note the derivation of $\neg A\:\vdash \: A\rightarrow B$ in 2-8. 
A: Yuck!  It looks like somebody is trying to give you a headache.
To solve this there are a couple of general tricks you'll need to implement.


*

*derive $\neg (C\lor D)\vdash \neg C$.

*derive $\neg(C\to D)\vdash C$.


Combining yields a derivation of $\neg(A\lor (A\to B))\vdash A\land\neg A$. 
Toward 1, after assuming $\neg (C\lor D)$ you'll want temporarily to assume $C$.  An application of $\lor$-intro gives you a contradiction.
Toward 2, you'll first want to have achieved


*derive $C,\neg C\vdash D$.


With 3 in hand, let's return to 2.  Assume $\neg (C\to D)$.  Temporarily assume $\neg C$.  Further temporarily assume $C$.  After reaching $D$ by trick 3, conditional proof gives $C\to D$.  You now have a contradiction, returning $\neg\neg C$.
Finally toward 3, the strategy is to 'convict the innocent': after assuming $C$ and $\neg C$, further temporarily assume $D$.  Here you can can assert $C\land \neg C$.  Using this proof of a contradiction which begins from an assumption of $D$ you can infer $\neg D$.
