Show the diagonal $\Delta \subseteq X\times X$ is diffeomorphic to $X$, so $\Delta$ is a manifold if $X$ is.
I wanted to prove this directly without using the notions of embeddings or the use of atlases. Anyway, what I have so far is (albeit I'm not sure if this is right):
Let $U$ be an open set of $X$, where $X$ is a $k$-dimensional manifold, and define the map $\psi:U\to\Delta$ by $u\mapsto(u,u)$. Then, $\forall u \in U$, there exists $\varphi : U \to \mathbb{R}^k$, where $\varphi$ is a diffeomorphism by definition. Now consider the set $$\psi(U) = \left\{ (x,x) \in \Delta \space\ \big| \space\ x \in U \right\}$$ Define the map $\phi: \psi(U)\to\mathbb{R}^k$ as $(u,u)\mapsto \varphi(u)$. Since $U$ is an arbitrary open set of $X$, it follows that for any $x \in X$, there exists some open $U$ such that $x \in U \subseteq X$, so $\psi$ is a diffeomorhpism.
The original problem doesn't make sense. How can you say "$X$ is diffeomorphic to $Y$ so $X$ is a manifold"? In order to talk about things being diffeomorphic you already have to know that they are manifolds.
So I assume this is what you want to prove: if $f:X\to M$ is a homeomorphism then we can introduce a differntial structure on $X$ making $f$ a diffeomorphism.
Indeed, let $U\subseteq M$ be open and $\varphi:U\to \mathbb{R}^n$ be a $C^k$ coordinate map. Let $V:=f^{-1}(Y)$ and define
$$\Phi: V\to \mathbb{R}^n$$ $$\Phi(x)=\varphi(f(x))$$
Now if $\{\Phi\}$ is a $C^k$-atlas on $X$ (which I leave as an exercise) then $f$ becomes $C^k$ differentiable. This follows simply because the composition
$$\mathbb{R}^{n}\xrightarrow{\Phi^{-1}} X\xrightarrow{f}M\xrightarrow{\phi}\mathbb{R}^m$$
is $C^k$ almost by definition.