Every finite-dimensional vector space V over $\mathbb{R}$ admits a smooth structure. Let $V$ be an $n$-dimensional vector space over $\mathbb{R}$. Consider the linear isomorphism $T:V\longrightarrow \mathbb{R}^{n}$. We define a topology in $V$. We say that $U\subseteq V$ is open in $V$ if $T(U)$ is open in $\mathbb{R}^{n}$. Clearly, if $U'\in \mathbb{R}^{n}$ is open, then $T^{-1}(U)$ is open in $V$. Also, $V$ and $\emptyset$ are open in $V$ since $T(V)=\mathbb{R}^{n}$ and $T(\emptyset)=\emptyset$ are open in $\mathbb{R}^{n}$. For an arbitrary collection $\{U_{i}:i\in I\}$ of open sets in $V$, then
$T\left(\bigcup_{i\in I}{U_{i}}\right)=\bigcup_{i\in I}{T(U_{i})},$
which is open in $\mathbb{R}^{n}$. And so $\bigcup_{i\in I}{U_{i}}$ is open in $V$. Similarly, for open sets $U_{1},\dotsc,U_{n}$ in $V$, $\bigcap_{i=1}^{n}{U_{i}}$ is also open in $V$ since $T\left(\bigcap_{i=1}^{n}{U_{i}}\right)=\bigcap_{i=1}^{n}{T(U_{i}})$ is open in $\mathbb{R}^{n}$. From the above observation, we can also observe that $T$ and $T^{-1}$ are continuous. Hence $T$ is a homeomorphism. But homeomorphism preserves Hausdorff property and second-countability of topological spaces. Therefore, $V$ is also Hausdorff and second-countable. Since $V$ is homeomorphic to $\mathbb{R}^{n}$, $V$ is an $n$-dimensional manifold.
To show that $V$ admits a smooth structure, note that $\{(T,V)\}$ is itself a smooth atlas. Therefore, it is contained in a maximal smooth atlas. Hence $V$ is a smooth manifold.
Is this reasoning correct?
 A: Your reasoning is correct. However, in my opinion you should additionally prove that toplogy and smooth structure do not depend on the choice of $T$.

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*It is clear that any bijection $b : V \to \mathbb R^n$ induces a unique topology on $V$ making $b$ a homeomorphism. However, it is not adequate to allow an arbitrary bijection. Your limitation to  linear isomorphism is perfect.
Note that allowing arbitrary bijections would prevent us to say that $V$ is an $n$-manifold. In fact, if there is a bijection $b : V \to \mathbb R^n$, then there bijections from $V$ to any $\mathbb R^m$.


*Let $T'$ be a second linear isomorphism. Then $S  = T' \circ T^{-1} : \mathbb R^n \to \mathbb R^n$ is also a linear isomorphism. Since all linear maps are continuous, we see that $S$ is a homeomorphims. But now we have
$T(U)$ open in $\mathbb R^n$ iff $S(T(U))$ open in $\mathbb R^n$ iff $T'(U)$ open in $\mathbb R^n$. This shows that $T, T'$ induce the same topology on $V$.


*For each linear isomorphism $T : V \to \mathbb R^n$ we get an atlas $\{(T, V)\}$ and the $n$-manifold $V$. It is automatically a smooth atlas because its only transition function is $id:  \mathbb R^n \to \mathbb R^n$ which is smooth. If $T'$ is a second linear isomorphism, then we may consider the atlas $\{(T, V), (T',V)\}$. It is a smooth atlas because its non-identity transition functions are $S =  T' \circ T^{-1}$ and $S^{-1}$. But all linear maps are smooth. This shows that $\{(T, V)\}$ and $\{(T',V)\}$ generate the same maximal smooth atlas.
