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I wish to understand the details of why a finite dimensional vector space is a topological manifold, particularly following Jonh Lee's Introduction to smooth manifolds. I know that this questions has been asked at least Here and here, and while I convinced myself that it is the case, I would not be able to fill in the details if challenged, hence I decided to open a new post with the relevant questions and approach.

Example 1.24 (Finite-Dimensional Vector Spaces). Let $V$ be a finite-dimensional real vector space. Any norm on $V$ determines a topology, which is independent of the choice of norm (Exercise B.49). With this topology, $V$ is a topological $n$-manifold, and has a natural smooth structure defined as follows. Each (ordered) basis $(E_1,...,E_n)$ for $V$ defines a basis isomorphism $E:R^n to V$ by
$$
E(x) = sum_{i=1}^nx^iE_i.
$$
This map is a homeomorphism [...].

I was filling the details and wanted to discuss my proofwriting when I was told that it was obvious because any real finite dimensional vector space is isomorphic to $mathbb{R}^n$. However at my stage I don't understand how that translates into showing each of the technical details in the definition of a topological manifold. Here are my questions:

How do I precisely show that $V$ is a topological manifold?

My reasoning is this: As stated, all norms are equivalent in a finite-dimensional vector space and generate the same topology, but which one is it? well, since a norm induces a metric, I thought it will be the metric topology (denote it by $tau_{||cdot||}$), i.e., the one generated by the basis $mathcal{B} = lbrace B_r(v) : uin V ; text{and}; r>0 rbrace$ where $B_r(u) = lbrace vin V : d_{||cdot||}(u,v) = ||u-v|| <rrbrace$, which would then be unique. Then $(V,tau_{||cdot||})$ would be my topological space. I can show that it is Hausdorff but I am not sure how to show it is second countable. any suggestion?

Then I need to show it is locally Eucliean of dimension $n$. For this, I need to show that for every point in $V$ there is a neighborood which is mapped to an open subset of $mathbb{R}^n$ (or $mathbb{R^n}$ itself). The given suggestion is the isomorphism $E$ (in linear algebra, a linear transformation which is also a bijection) which the author claims is a homeomorphism. I am guessing it means precisely a homeomorphism from $E: (V,tau_{||cdot||}) to (mathbb{R}^n,tau_U)$ where $tau_U$ is the usual topology. So we would be showing that $V$ is homeomorphic to $R^n$, which implies $V$ is locally Euclidean, is this correct? Now I already now that $E$ is a biyection and since $mathcal{B}$ is a basis for $tau_{||cdot||}$, I could show that $E(mathcal{B})$ is a basis for $tau_U$, which should be duable since $n$-balls form a basis for $tau_U$ as well, and I think I would be done. Is that right? Is this a good way of proceeding? I will probably also have questions regarding on how to write these ideas, but one step at the time. Unfortuately I am learning all of this at the same time as opposed to a typical math degree curriculum, so I appretiate your patience.

If you show that $V$ with the metric topology induced by any norm is homeomorphic to $Bbb R^n$ using the suggested vector space isomorphism, then since $Bbb R^n$ is Hausdorff, second countable, and locally Euclidean, the same things apply to $V$, so you will have automatically verified all three properties of what it means to show that $V$ is a topological manifold.