I added $\forall Y(\emptyset \cup Y = Y)$ as a premise; the exercise does not provide it.
$ \def\fitch#1#2{\quad\begin{array}{|l}#1\\\hline#2\end{array}} \def\Ae#1{\qquad\mathbf{\forall E} \: #1 \\} \def\Ai#1{\qquad\mathbf{\forall I} \: #1 \\} \def\Ee#1{\qquad\mathbf{\exists E} \: #1 \\} \def\Ei#1{\qquad\mathbf{\exists I} \: #1 \\} \def\R#1{\qquad\mathbf{R} \: #1 \\} \def\ci#1{\qquad\mathbf{\land I} \: #1 \\} \def\ce#1{\qquad\mathbf{\land E} \: #1 \\} \def\ii#1{\qquad\mathbf{\to I} \: #1 \\} \def\ie#1{\qquad\mathbf{\to E} \: #1 \\} \def\be#1{\qquad\mathbf{\leftrightarrow E} \: #1 \\} \def\bi#1{\qquad\mathbf{\leftrightarrow I} \: #1 \\} \def\qi#1{\qquad\mathbf{=I}\\} \def\qe#1{\qquad\mathbf{=E} \: #1 \\} \def\ne#1{\qquad\mathbf{\neg E} \: #1 \\} \def\ni#1{\qquad\mathbf{\neg I} \: #1 \\} \def\IP#1{\qquad\mathbf{IP} \: #1 \\} \def\x#1{\qquad\mathbf{X} \: #1 \\} \def\DNE#1{\qquad\mathbf{DNE} \: #1 \\} $ $ \fitch{1.\, \forall Y(\emptyset \cup Y = Y)}{ 2.\, \exists X\forall Y(X \cup Y = Y) \Ei{1} \fitch{3.\, \forall Y(C \cup Y=Y)}{ \fitch{4.\, \forall Y(D \cup Y=Y)}{ 5.\, C \cup D=D \Ae{3} 6.\, D \cup C = C \Ae{4} \vdots\\ C = D }\\ \forall Y(D \cup Y=Y) \to C=D\\ \forall Z(\forall Y(Z \cup Y=Y) \to C=Z)\\ k. \forall Y(C \cup Y=Y) \land \forall Z(\forall Y(Z \cup Y=Y) \to C=Z) \ci{4,k} k+1.\, \exists X[\forall Y(X \cup Y=Y) \land \forall Z(\forall Y(Z \cup Y=Y) \to X=Z] \Ei{k+1} }\\ \exists X[\forall Y(X \cup Y=Y) \land \forall Z(\forall Y(Z \cup Y=Y) \to X=Z] \Ee{3-k+1} } $
Is this proof skeleton correct ? Am I going in a good direction ?
EDIT:
Following advice in the comments, I made another version including 3 axioms as premises and deriving $\forall Y(\emptyset \cup Y = Y)$. I omitted the instantiation of the axioms. Going to include it in a final version.
$ \fitch{1.\, \forall x\neg(x \in \emptyset)\\2.\,\forall A\forall B \forall x(x \in A \cup B \leftrightarrow a \in A \lor a \in B)\\3.\,\forall A\forall B(\forall x(x \in A \leftrightarrow x \in B) \to A=B)}{ \fitch{4.\, a \in \emptyset \cup A}{ 5.\, a \in \emptyset \lor a \in A \\ 6.\, \neg(a \in \emptyset)\\ 7.\,a \in A }\\ \fitch{8.\, a \in A}{ 9.\, a \in A\\ 10.\,a \in \emptyset \lor a \in A\\ 11.\,a \in \emptyset \cup A }\\ 12.\, a \in \emptyset \cup A \leftrightarrow a \in A\\ 13.\,\, \forall Y(\emptyset \cup Y=\emptyset) \\ 14.\, \exists X\forall Y(X \cup Y=X) \\ \fitch{\forall Y(C \cup Y=Y)}{ \fitch{4.\, \forall Y(D \cup Y=Y)}{ 5.\, C \cup D=D \Ae{3} 6.\, D \cup C = C \Ae{4} \vdots\\ C = D }\\ \forall Y(D \cup Y=Y) \to C=D\\ \forall Z(\forall Y(Z \cup Y=Y) \to C=Z)\\ k. \forall Y(C \cup Y=Y) \land \forall Z(\forall Y(Z \cup Y=Y) \to C=Z) k+1.\, \exists X[\forall Y(X \cup Y=Y) \land \forall Z(\forall Y(Z \cup Y=Y) \to X=Z] \Ei{} }\\ \exists X[\forall Y(X \cup Y=Y) \land \forall Z(\forall Y(Z \cup Y=Y) \to X=Z] \Ee{} } $
P.S.: It is Example 3.6.2 of the book "How to Prove It" by Daniel Velleman.