I am trying to understand the proof of this statement.
Any Hausdorff topology on a finite-dimensional vector space with respect to which vector space operations are continuous is equivalent to the usual one.
Source: https://personal.math.ubc.ca/~cass/research/pdf/TVS.pdf
In the proof it is stated that:
It remains to show that the inverse of $f$ is continuous. For this, it suffices to show that $f(B(1))$ contains a neighbourhood of $0$.
My Question is: Why does it suffice to show that $f(B(1))$ contains a neighborhood of $0$?
Here the proposition in detail:
Any Hausdorff topology on a finite-dimensional vector space is equivalent to the usual one
Proof: A basis of $V$ determines a linear isomorophism $f:\mathbb{R}^n \rightarrow V$, which is continuous by the assumptions on the topology of $V$. By definition of continuity, if $U$ is any neighborhood of $0$ in $V$ there exists some disk $B(r)$ with $f(B(r))\subseteq U$.
It remains to show that the inverse of $f$ is continuous. For this, it suffices to show that $f(B(1))$ contains a neighborhood of $0$. . . .
It is well-known that a linear map $\phi : V \to W$ between TVS $V,W$ is continuous iff it is continuous at $0$. I shall give a proof later.
To prove that $g := f^{-1} : V \to \mathbb R^n$ is continuous, it therefore suffices to show that for each open neighborhood $W$ of $0$ in $\mathbb R^n$ there exists an open neighborhood $W'$ of $0$ in $V$ such that $g(W') \subset W$. The latter is equivalent to $W' \subset f(W)$.
Assume that $f(B(1))$ contains an open neighborhood $W_0$ of $0$ in $V$. There exists $r > 0$ such that $B(r) \subset W$. Then $W' := rW_0$ is an open neighborhood of $0$ in $V$ such that $$W' = rW_0 \subset r f(B(1)) = f(rB(1)) =f(B(r)) \subset f(W) .$$
Let us finally prove that a linear $\phi : V \to W$ which is continuous at $0$ is continuous at all $v \in V$.
Let $W'$ be an open neighborhood of $\phi(v)$ in $W$. Then $W'' = W' - \phi(v)$ is an open neighborhood of $0$ in $W$ and there exists an open neighborhood $V''$ of $0$ in $V$ such that $\phi(V'') \subset W''$. Clearly $V' = V'' + v$ is an open neighborhood of $v$ in $V$ and we have $$\phi(V') = \phi(V'' + v) = \phi(V'') + \phi(v) \subset W'' + \phi(v) = W'.$$