Let $X$ be a conected topological $n$-manifold, and $f:X\rightarrow X$ an homeomorphism, the Mapping Torus $M_f$ is defined as, $$M_f=X\times [0,1]/\sim$$ where $(x,0)\sim (f(x),1)$.
I am trying to prove that this is a manifold. Here is my approach:
This is obvious for every point $(x,a)$ with $a\neq 0,1$, so let's call $p\in M_f$ to $p=\pi(x_0,0)=\pi(f(x_0),1)$ where $\pi$ is the canonical projection. Let $\{\varphi, U\}$ be a chart of $X$ at $x_0$, and define,
$$\begin{matrix} V_1&=&U\times [0,1/2)\\ V_2&=&U\times (1/2,1] \end{matrix}$$
we have that, if $V=V_1\cup V_2$, $\pi(V)$ is an open set in $M_f$ and if we define, $$\begin{matrix} \phi:&V&\longrightarrow &\mathbb{R}^n\times (-1/2,1/2)\\ &(x,a)&\longmapsto&\left\{ \begin{matrix}(\varphi(x),a)&\text{if}&(x,a)\in V_1\\((\varphi\circ f^{-1})(x),a-1)&\text{if}&(x,a)\in V_2\end{matrix}\right. \end{matrix}$$ from here I think it should be enough to show that $\phi$ is a quotient map, since the only points with same image are those related by $\sim$.
Is this argument valid?, and also is this the correct way to solve this problem or is there anything more elegant?
You have the right idea, and this will work. There's nothing particularly more elegant, especially if (like me) you enjoy the elegance of the quotient topology. However, there are a couple of mistakes.
First is a typo: the formulas for $V_1$ and $V_2$ should have Cartesian product symbols, not union symbols.
Second, the formula for $V_2$ is wrong. It should be $$V_2 = f(U) \times (1/2,1] $$