Proof clarification request: boundedness of operator (related to finite-dimensional vector spaces). I have actually asked this question on MSE (here), and the answer given used the compactness of the unit ball. After doing some search, I found this question (it looks like mine is a duplicate of this one) and this (apparently) doesn't use the compactness  of the unit ball.
Context. On trying to prove that every finite-dimensional normed space is topologial isomorphic to $\Bbb R^n$, we define a generic normed space $E = (E,\|\cdot\|_E)$ such that $\dim(E)=n \in \Bbb N.$ Define
$$
T:\mathbb{R}^n\to E\\
x\mapsto x_1\delta_1+\dots x_n\delta_n
$$
where $B_E=\{\delta_1,\dots,\delta_n\}$ is an algebraic basis for $E$.
Although, the proof provided here isn't very clear (or detailed) to me (I am new to functional analysis). I am specifically talking about the part where the OP proves that $T^{-1}$ is bounded (or continuous).
Step $\mathbf{1.}$ Assuming $T^{-1}$ is not bounded: here, the OP states:

Suppose by contradiction that $T^{-1}$ is not bounded. Then, there exists
$$
(y_n)_n\subset E:\qquad \frac{\|T^{-1}y_n\|_{\mathbb{R}^n}}{\|y_n\|_{E}}>n\,\,\,\,\,\,\,\,\,\,\forall n\in\mathbb{N}
$$

And I can't figure how one would come up with such a sequence. My line of thought is as follows: If $T^{-1}$ is not bounded, then
$$ \forall c>0, \exists v \in E, \quad \|T^{-1}v\|_{\Bbb R^n} > c\|v\|_E $$
I understand that, in particular, we can pick $c=n$ and it follows that
$$ \exists v \in E\, : \, \|T^{-1}v\|_{\Bbb R^n} > n\|v\|_E \Leftrightarrow \frac{\|T^{-1}v\|_{\Bbb R^n}}{\|v\|_E} > n, \quad \text{ for $v \neq 0_E.$} $$
Well, obviously here $v \neq 0_E$ isn't a problem since $\|T^{-1}0_E\|_{\Bbb R^n} = 0$ which doesn't satisfy the strict inequality.
Now, how does the OP comes up with a infinite sequence? As far as I am concerned, it could be only one point $v \in E$ that satisfies the inequality... Basically, how is the sequence $(y_n)_n$ defined here?
Step $\mathbf{2.}$ After this, the OP defines a new sequence just like I show below:

Consider
$$
x_n:=\frac{T^{-1}y_n}{\|T^{-1}y_n\|_{E}},\quad (x_n)_n\subset\mathbb{R}^n
$$
We have
$$
\|x_n\|_{\mathbb{R}^n}=1\quad\forall n\in\mathbb{N}
$$

I have no problem understanding this part (I believe the OP made a mistake and it should be $\|T^{-1}y_n\|_{\Bbb R^n}$ in the denominator).
Step $\mathbf{3.}$ The next step in the proof from the OP is presented below.

but
$$
\|Tx_n\|_{E}<\frac{1}{n}\quad\forall n
$$
This is a contradiction since $T$ is injective.

Here, I understand that, since the sequence $(x_n)_n$ is bounded $(\|x_n\|_{\Bbb R^n} = 1 \Rightarrow \|x_n\|_{\Bbb R^n} \leqslant 1)$, then by the Bolzano Weierstrass Theorem there is a convergent subsequence to some element $x$. But, from here, I don't know how to reproduce the inequality above.
Thanks for any help in advance.
 A: In the interests of not answering in the comments...

*

*Really, you are trying to show any finite dimensional vector space $E$ is (linearly) isomorphic to $\Bbb R^{\dim E}$. This is stronger than, and more important than, mere homeomorphism. I use $\dim E$ so as not to overload the letter "$n$", which you did in your post

*If such $v$ exist for all $n$, you can just let $y_n$ be such a $v$ for $n=1,2,\cdots$, there is no problem there. It is impossible that $(y_n)$ is (eventually) constant, since there are always $n>\|v\|$ (this is the Archimedean property mentioned in the comments) but it doesn't matter at all for the proof. You just want a bag of vectors with arbitrarily large $T^{-1}$-image: the size of the bag (how many distinct elements) is irrelevant, though necessarily it is countably infinite

*$\|T^{-1}y_n\|$ is nonzero by definition, for all $n$, so the vector: $$x_n:=\frac{T^{-1}y_n}{\|T^{-1}y_n\|}$$Is well defined. Note: $$Tx_n=\frac{TT^{-1}y_n}{\|T^{-1}y_n\|}=\frac{y_n}{\|T^{-1}y_n\|}$$So: $$\|Tx_n\|=\frac{\|y_n\|}{\|T^{-1}y_n\|}<\frac{1}{n}$$By taking reciprocals for the definition inequality for the $(y_n)$

*The unit sphere of $\Bbb R^{\dim E}$, $S$, wherein the $(x_n)$ reside, is compact. If you already know $T$ is continuous, then $TS$ is known to be compact and, in particular, closed in $E$ ($E$ has a metric this Hausdorff topology). As the $Tx_n\overset{n\to\infty}{\longrightarrow}0$ in $TS$, $0$ is an element of $TS$. But that implies $T\alpha=0$ for some $\alpha\in S$, and as $\|\alpha\|=1\neq0$, $\alpha\neq0$ and $T$ has a nontrivial kernel. It follows that $T$ is not injective, which is a contradiction. It's late and I'm tired, so I can't think of a non-topological way to argue this right now (I think that's what you want), sorry

*Update: Your Bolzano-Weierstrass mention was apt. There is an accumulation point $\alpha$ of $(x_n)$, and if you already know $T$ is continuous you can say: $Tx_{n_k}\to T\alpha$ along some subsequence, but the $Tx_n\to0$ thus $T\alpha=0$ since limits are unique (the underlying topologies are $\rm T1$). This contradicts the injectivity of $T$. Note: this is topological in secret, since we still use continuity ideas and the Bolzano-Weierstrass theorem (and the Heine-Borel theorem) are just shadows of the general theorem: "A metric space (e.g. $S$) is sequentially compact iff. it is compact iff. it is complete and totally bounded." You can verify $S$ is compact via the last criterion

*That every linear map out of a finite dimensional linear space is continuous is much more easily seen (in my opinion) by crudely bounding the matrix entries. This argument involves essentially no topology, and broadly goes like this: fix a $T:U\to V$ where $U$ has dimension $n$ and $V$ is an arbitrary vector space (real, but may have infinite dimension). Fix an orthonormal vector basis $(e_i)_{i=1}^n$, for $U,V$ respectively. By finitude, there is a finite constant $K=\max(\|Te_i\|;i=1,2,\cdots,n)$. Any $x$ in the unit sphere $S$ of $U$ is expressible as $\sum_{i=1}^n\lambda_i e_i$ for some scalars $(\lambda_i)$ - $\|Tx\|\le K\sum_{i=1}^n|\lambda_i|$ follows from the triangle inequality and the definition of $K$. Note that: $$1=\|\lambda_1e_1+\cdots+\lambda_ne_n\|=\sum_{i=1}^n\lambda_i^2$$Which is just the general Pythagorean identity (I picked $(e_i)$ orthonormal... e.g. appeal to Gram-Schmidt). Clearly, each $\lambda_i$ is less than or equal to $1$ in magnitude - so: $$\|Tx\|\le nK$$You can do better and bound by $\sqrt{n}$ from Cauchy-Schwarz. This is also very similar to how you prove any two norms are equivalent in finite dimensions.

