$∀x(P(x)∨¬P(x))$ fitch proof without premises? How should I go about solving this? Am I able to solve this with contradiction? 
I tried starting with $¬∀x(P(x)∨¬P(x))$, but I don't know where to go with it. 
Some help would be nice, thank you
Yes, you can prove this by contradiction, but not assume $¬∀x(P(x)∨¬P(x)),$ instead we will assume $¬(P(a)∨¬P(a))$ so we can use $\forall$ Intro :$$\def\fitch#1#2{\quad\begin{array}{|l}#1\\\hline#2\end{array}}
\fitch{1.}{\fitch{2.~~\boxed{a}}{\fitch{3.~~\neg(P(a)\lor\neg P(a))}{\fitch{4.~~P(a)}{5.~~P(a)\lor\neg P(a)\hspace{10ex}\lor\text{ Intro }4\\6.~~\bot\hspace{22.9ex}\bot\text{ Intro }3,5}\\7.~~\neg P(a)\hspace{22ex}\neg\text{ Intro }4-6\\8.P(a)\lor\neg P(a)\hspace{14.5ex}\lor\text{ Intro }7\\9.~~\bot\hspace{26.2ex}\bot\text{ Intro }3,8}\\10.~~P(a)\lor\neg P(a)\hspace{16.9ex}\neg\text{ Intro }3-9}\\11.~~\forall x(P(x)\lor\neg P(x))\hspace{16ex}\forall\text{ Intro }2-10}$$
Yes, you can prove this by contradiction, but not assume $¬∀x(P(x)∨¬P(x)),$ instead we will assume $¬(P(a)∨¬P(a))$ so we can use $\forall$ Intro :$$\def\fitch#1#2{\quad\begin{array}{|l}#1\\\hline#2\end{array}}
\fitch{1.}{\fitch{2.~~\boxed{a}}{\fitch{3.~~\neg(P(a)\lor\neg P(a))}{\fitch{4.~~P(a)}{5.~~P(a)\lor\neg P(a)\hspace{10ex}\lor\text{ Intro }4\\6.~~\bot\hspace{22.9ex}\bot\text{ Intro }3,5}\\7.~~\neg P(a)\hspace{22ex}\neg\text{ Intro }4-6\\8.P(a)\lor\neg P(a)\hspace{14.5ex}\lor\text{ Intro }7\\9.~~\bot\hspace{26.2ex}\bot\text{ Intro }3,8}\\10.~~P(a)\lor\neg P(a)\hspace{16.9ex}\neg\text{ Intro }3-9}\\11.~~\forall x(P(x)\lor\neg P(x))\hspace{16ex}\forall\text{ Intro }2-10}$$