Let $(V,\|\cdot \|)$ a normed space of dimension $n$. I want to prove rigorously that it's a smooth manifold. This is how I do:
$(V,\|\cdot \|)$ is a topological manifold :
Let $x\in V$, $(v_1,...,v_n)$ a basis of $V$ and
\begin{align*}
L:\mathbb R^n&\longrightarrow V\\
(x_1,...,x_n)&\longmapsto \sum_{i=1}^n x_iv_i.
\end{align*}
$L$ is obviously continuous (for the topology induced by $\|\cdot \|$). Moreover $L$ is bijective since the fact that $v_1,...,v_n$ is a basis implies that $$L(x_1,...,x_n)=0\implies (x_1,...,x_n)=0$$ and thus $\ker L=\{0\}$, therefore it's injective and thus bijective. We know that the inverse of a linear map is also linear, and thus $L^{-1}$ is also continuous (since it's continuous at $0$).
Therefore $L$ is an homeomorphism. and thus $(V,L^{-1})$ is a chart of $V$, and thus $\mathcal A=\{(V,L^{-1})\}$ is an atlas. Therefore it's a topological manifold.
$(V,\|\cdot \|)$ is a smooth manifold.
I have to show that for each chart pair $(V,L^{-1})$ and $(V,T^{-1})$ there is a diffeomorphism. I know that if $(w_1,...,w_n)$ is an other basis, there is an invertible matrix $(A_{ij})$ s.t. $v_i=\sum_{j=1}^n A_{ij}w_j$. How can I continue ?
Well, if you take an atlas with only one chart $L^{-1}$ for $V$ (and enlarge it to a maximal atlas), you don't have to check anything. If you want to show that if you choose a different basis and obtain a different map $T \colon \mathbb{R}^n \rightarrow V$ then the map $T^{-1}$ is smooth (or, in order words, that $T^{-1}$ belongs to the maximal atlas generated by $L^{-1}$), you indeed need to check that $T^{-1} \circ L \colon \mathbb{R}^n \rightarrow \mathbb{R}^n$ is smooth as a map from $\mathbb{R^n}$ to $\mathbb{R}^n$. But this is a linear map, and so of course it is smooth.