A question about atlas in smooth differentiable I know this is definitely an easy question. However, I just cannot figure out.

I am using Do Carmo's Riemannian Geometry to teach myself Differential Geometry.

In his book, he gave a definition of differentiable manifold:

A differentiable manifold of dimension $n$ is a set $M$ and a family of injective mappings $x_\alpha:U_\alpha\subset \mathbb{R}^n\to M$ of open sets $U_\alpha$ of $\mathbb{R}^n$ into $M$ such that:

(1)$\bigcup_\alpha x_\alpha(U_\alpha)=M$

(2)for any pair $\alpha,\beta$, with $x_\alpha(U_\alpha)\cap x_\beta(U_\beta)=W\neq\emptyset$, the sets $x_\alpha^{-1}(W)$ and $x_\beta^{-1}(W)$ are open sets in $\mathbb{R}^n$ and the mappings $x_\beta^{-1}\circ x_\alpha$ are differentiable.

(3) The family $\{(U_\alpha,x_\alpha)\}$ is maximal relative to the conditions (1) and (2).

Furtuhermore, he define $A\subset M$ to be an open set in $M$ iff $x_\alpha^{-1}\left(A\cap x_\alpha(U_\alpha)\right)$ is an open set in $\mathbb{R}^n$ for every $\alpha$.

I can show that this is a topology, and that $x_{\alpha}(U_\alpha)$ is open and $x_\alpha$ is continuous. However, I have no idea how to prove that $x_\alpha$ is actually a homeomorphism. It suffices to prove that $x_\alpha$ is a open mapping, but I have no idea.

Any help will be appreciated. Thanks.
We check that $x_{\alpha}:U_{\alpha}\rightarrow M$ is an open mapping. So given $U\subset U_{\alpha}$ open, we only need to see that $x_{\alpha}(U)\subset M$ is open according to the definition you gave. 

For any $\beta$, we have 
\begin{align}x_{\beta}^{-1}\left(x_{\alpha}(U)\cap x_{\beta}(U_{\beta})\right)&amp;=(x_{\beta}^{-1}\circ x_{\alpha})(U)\cap (x_{\beta}^{-1}\circ x_{\beta})(U_{\beta})=(x_{\alpha}^{-1}\circ x_{\beta})^{-1}(U)\;\cap\; U_{\beta}\end{align}

which is  open because $x_{\alpha}^{-1}\circ x_{\beta}$ is differentiable.
Hint: The $x_\alpha$ are quotient maps by definition, but they are also injective.