I was reading this post, but unfortunately I don't understand the following statement made in the following answer given by Amitai Yuval:
"About the topology: Every $(U_\alpha,\varphi_\alpha)$ needs to be a chart, meaning that $\varphi_\alpha$ is a homeomorphism on its image, so for every subset $V\subset U_\alpha$, $V$ is open if and only if $\varphi_\alpha(V)\subset\mathbb{R}^n$ is open. Since the $U_\alpha$s cover $M$, it means that the collection of open subsets of elements in $\{\varphi_\alpha(U_\alpha)\}$ induces a basis of a topology on $M$, thus the topology is indeed unique."
Denote with $\mathcal{T}_\mathcal{B}$ the topology induced by $\mathcal{B}$ e we suppose that for all $\alpha$, the pair $(U_\alpha,\varphi_\alpha)$ are chart for $M$, then on $M$ exists a topology $\mathcal{T}$ such that $U_\alpha\in\mathcal{T}$ and $\varphi_\alpha\colon U_\alpha\to\varphi_\alpha(U_\alpha)$ is an homeomorphism.
Claim: $\mathcal{T}_\mathcal{B}=\mathcal{T}.$
First, for each charts we denote with $\mathcal{T}_{\mathcal{B}_{U\alpha}}$ the topology induced by $\mathcal{T}_\mathcal{B}$ on $U_\alpha$, and with $\mathcal{T}_{U_\alpha}$ the topology induce by $\mathcal{T}$ on $U_\alpha$.
Let $A\in\mathcal{T}_\mathcal{B}$, since $U_\alpha\in\mathcal{T}_B$ for all $\alpha$, $A\cap U_\alpha\in\mathcal{T}_\mathcal{B}$, then $A\cap\ U_\alpha=(A\cap U_\alpha)\cap U_\alpha\in\mathcal{T}_{\mathcal{B}_{U_\alpha}}.$
Since $\varphi_\alpha\colon (U_\alpha,\mathcal{T}_{\mathcal{B}_{U_\alpha}})\to\varphi_\alpha(U_\alpha)$ is an homeomorphism , $\varphi_\alpha(A\cap U_\alpha)\subseteq\varphi_\alpha(U_\alpha)$ is open. On the other hand, also
$\varphi_\alpha\colon (U_\alpha,\mathcal{T}_{U_\alpha})\to\varphi_\alpha(U_\alpha)$ is an homeomorphism, then we have that $$\varphi_\alpha^{-1}(\varphi_\alpha(A\cap U_\alpha))=A\cap U_\alpha\in\mathcal{T}_{U_\alpha}\subseteq\mathcal{T}.$$
Since $$A=\bigcup_\alpha (A\cap U_\alpha)\Rightarrow A\in\mathcal{T}.$$
The above shows that $\mathcal{T}_{\mathcal{B}}\subseteq\mathcal{T}.$ In the same way the vice versa can be shown.
It's correct?
Question. What is explicitly meant in this answer, in which we want to show the uniqueness of the topology that satisfies the hypothesis of the lemma?
Thanks!
What "each $(U_\alpha,\varphi_\alpha)$ needs to be a chart" means is "the topology on $M$ needs to be such that, for any $\alpha$, a subset $V \subseteq U_\alpha$ is open if and only if $\varphi_\alpha(V)$ is open in $\mathbb{R}^n$". The only point that Amitai Yuval was making is that "for any $\alpha$, a subset $V \subseteq U_\alpha$ is open if and only if $\varphi_\alpha(V)$ is open in $\mathbb{R}^n$" uniquely determines a topology on $M$, namely, the topology in which a subset $V$ of $M$ is open if and only if $\varphi_\alpha(V\cap U_\alpha)$ is open in $\mathbb{R}^n$ for each $\alpha$.
Side remark: you should have commented this question on his answer rather than made it a new question on its own.