I am working on this proof and I have set it up. I'm not sure how I should proceed or if I am doing it correctly so far Thank you!
The first 3 lines before the horizontal line are the premises!

I am working on this proof and I have set it up. I'm not sure how I should proceed or if I am doing it correctly so far Thank you!
The first 3 lines before the horizontal line are the premises!

On
$\def\fitch#1#2{\begin{array}{|l}#1\\\hline #2\end{array}}\def\magic{\operatorname{Magical}}\def\mythic{\operatorname{Mythical}}$ Your basic idea is okay, but you cannot use $\vee$-elimination unless you have a disjunction; that is $\mythic(c)\vee\neg\mythic(c)$
If you cannot use the Law of Excluded Middle to assert this, then you can modify the proof to use another negation introduction.
$$\small\fitch{~\vdots}{\fitch{\neg\magic(c)}{\fitch{\mythic(c)}{~\vdots\\ \bot}\\\fitch{\neg\mythic(c)}{~\vdots\\ \bot}\\\color{red}\bot\quad\color{red}{\vee\mathsf E\text{ invalid}}}\\\neg\neg \magic(c)\quad\neg\mathsf I\\\magic(c)\qquad\neg\neg\mathsf E}\qquad\qquad\fitch{~\vdots}{\fitch{\neg\magic(c)}{\fitch{\mythic(c)}{~\vdots\\ \bot}\\\neg\mythic(c)\qquad\neg\mathsf I\\~\vdots\\ \bot}\\\neg\neg\magic(c)\quad\neg\mathsf I\\ \magic(c)\qquad\neg\neg\mathsf E}$$
On
Here is an easy way to fix the proof: keep the subproof that assumes $Mythical(c)$ and ends with $\bot$. Close this subproof using $\neg \ Intro$ to get $\neg Mythical(c)$, and now you can do all the steps you originally did inside the subproof that assumed $\neg Mythical(c)$ to get to $\bot$, as desired. In other words, there is no need for two separate subproofs, and hence no need to prove the Law of Excluded Middle either.
Follows your set up, here is another approach:
$$\boxed{\begin{align} &\text{My(c):=Mythical(c)}\\ &\text{Ma(c):=Mammal(c)}\\ &\text{Mo(c):=Mortal(c)}\\ &\text{Ho(c):=Horned(c)}\\ &\text{Mg(c):=Magical(c)}\\ \end{align}}$$
Here is how to continue with what you have and finish the proof use $\lor$ Elim:
That proved $My(c)\lor\neg My(c)$ now we can use $\lor$ Elim.
Which will take a little more works.