Proposition Let $Y$ be a quasi-projective variety, then $Y$ is covered by the open sets $Y\cap U_i$, $i=0,\dots, n$, which are homeomorphic to quasi-affine varieties via the mapping $\varphi_i\colon U_i\to \mathbb{A}_k^n$ defined as: if $P=(a_0,\dots, a_n)\in U_i$, then $\varphi(P)=Q$, where $Q$ is the point with affine coordinates $$\bigg(\frac{a_0}{a_i},\cdots,\frac{a_{i-1}}{a_{i}},\frac{a_{i+1}}{a_{i}},\cdots ,\frac{a_n}{a_i}\bigg)$$
proof. Let $Y$ be a quasi-projective variety and consider $Y\cap U_i$.
Question. $Y\cap U_i$ is open in $U_i$?
If the answer to the above question is yes, then since $\varphi_i\colon U_i\to \mathbb{A}_k^n$ it is a homeomorphism, $\varphi_i(Y\cap U_i)$ is open in $\mathbb{A}_k^n$, therefore $\varphi_i(Y\cap U_i)$ is irreducible and dense in $\mathbb{A}_k^n$, that is $$\overline{\varphi_i(Y\cap U_i)}=\mathbb{A}_k^n,$$ hence $$\varphi_i(Y\cap U_i)\subset\overline{\varphi_i(Y\cap U_i)}$$ is a quasi projective variety.
Finally, since the restriction of a homomorphism is a homeomorphism, we have that $$Y\cap U_i \cong \varphi_i(Y\cap U_i)$$
Edit 1. In fact perhaps, there is no need to observe that $\varphi(Y\cap U_i)$ is dense and irreducible in $\mathbb{A}_k^n$, because $\mathbb{A}_k^n$ is an affine variety and therefore $\varphi(Y\cap U_i)$ is an open set in a affine variety, that is quasi-affine variety by definition.
Edit 2 Thanks to KReiser's comment and thanks to him2020's detailed answer I realized that $Y\cap U_i$ is not open in $U_i$, at this point, however, my proof of the above proposition is unsuccessful. Since I know that the thesis of this proposition is true how can I show it at this point since my procedure is wrong?
Thanks!
Best Answer
Question: "Is $Y∩U_i$ open in $U_i$ for all $i=0,..,n$?"
Answer: Let $Y \subseteq \mathbb{P}^n_k$ be quasi projective over $k$, where $k$ is the field of complex numbers. Assume $Y:=V(I)$ where $I \subseteq A[x_0,..,x_n]$ is a homogeneous prime ideal, and where $dim(Y):=k < n$. It follows $V_i:=U_i \cap Y \subseteq U_i$ is an open subscheme of $Y$ and hence $dim(V_i) \leq dim(Y)=k < n$ hence $V_i$ cannot be open in $U_i$. An open subscheme $U \subseteq U_i \cong \mathbb{A}^n_k$ must have $dim(U)=n=dim(U_i)$. Hence the answer to your question is: Not in general.
Edit 2: "Thanks to KReiser's comment and thanks to him2020's detailed answer I realized that Y∩Ui is not open in $U_i$ , at this point, however, my proof of the above proposition is unsuccessful. Since I know that the thesis of this proposition is true how can I show it at this point since my procedure is wrong? Thanks!"
Your claim: "..therefore $φ_i(Y∩U_i)$ is irreducible and dense in $A^n_k$,"
Answer: Since $dim(Y \cap U_i)< n$ it cannot be dense in $U_i\cong \mathbb{A}^n_k$ which has dimension $n$.
Comment: "Thanks for the answer, but maybe I didn't make myself clear, how can I prove the above proposition? – Nat. 17 hours ago"
Proposition. "Let Y be a quasi-projective variety, then Y is covered by the open sets $Y∩U_i$, i=0,…,n, which are homeomorphic to quasi-affine varieties via the mapping $φ_i:U_i→A^n_k$."
Proof: Since $Y \subseteq \mathbb{P}^n_k$ is quasi projective, there is a projective variety $Y' \subseteq \mathbb{P}^n_k$ with $Y \subseteq Y'$ an open subvariety. It follows $Y' \cap U_i \cong \mathbb{A}^n_k$ is an affine algebraic variety, and $Y \cap U_i \subseteq Y' \cap U_i$ is an open subvariety of an affine variety, hence $V_i:=Y\cap U_i$ is a quasi affine variety.QED.