With any bijection between two sets, if one of the sets carries a topology, you can transport the topology to the other. The bijection then becomes a homeomorphism.
For topological vector spaces, of course not every bijection will transport a TVS topology on the one space to a TVS topology on the other, but a linear isomorphism will (elementary but tedious verifications left as an exercise).
So given any finite-dimensional $\mathbb{K}$ vector space $V$, with any linear isomorphism, we can transport its topology to $\mathbb{K}^n$.
The claim is: If the topology $\mathcal{T}$ on $V$ was a Hausdorff TVS topology, then the topology $\mathcal{T}_V$ on $\mathbb{K}^n$ induced by the linear isomorphism is the standard (product) topology $\mathcal{S}_n$ on $\mathbb{K}^n$.
Let $(e_i)_{1 \leqslant i \leqslant n}$ the standard basis in $\mathbb{K}^n$.
First, we note that the map $\lambda \mapsto \lambda\cdot e_i$ is continuous as a map $\bigl(\mathbb{K},\,\mathcal{S}_1\bigr) \to \bigl(\mathbb{K}^n,\, \mathcal{T}_V\bigr)$, since $\mathcal{T}_V$ is a TVS topology.
Hence the map $\mathbb{K}^n \to \bigl(\mathbb{K}^n\bigr)^n$ given by $(\lambda_1,\, \dotsc,\,\lambda_n) \mapsto (\lambda_1\cdot e_1,\, \dotsc,\, \lambda_n\cdot e_n)$ is continuous for the product topologies obtained from $\mathcal{S}_1$ resp. $\mathcal{T}_V$.
Now, $\mathcal{T}_V$ being a TVS topology, addition is continuous, therefore
$$\begin{gather}
\Phi \colon \bigl(\mathbb{K}^n,\, \mathcal{S}_n\bigr) \to \bigl(\mathbb{K}^n,\, \mathcal{T}_V\bigr)\\
(\lambda_1,\, \dotsc,\, \lambda_n) \mapsto \sum_{i = 1}^n \lambda_i \cdot e_i = (\lambda_1,\, \dotsc,\, \lambda_n)
\end{gather}$$
is continuous ($\mathcal{S}_n$ is the product topology of $\mathcal{S}_1$).
Therefore:
The standard topology is the finest TVS topology that $\mathbb{K}^n$ can carry.
Note that we did not assume the Hausdorff property, so that result holds even if Hausdorffness is not required.
Next, we remark
In a topological $\mathbb{K}$ vector space $E$, $0$ has a neighbourhood basis consisting of balanced sets.
(A set $B \subset E$ is balanced if $(\forall \lambda \in \mathbb{K})(\lvert \lambda\rvert \leqslant 1 \Rightarrow \lambda\cdot B \subset B)$.)
That follows from the continuity of scalar multiplication $\mathbb{K} \times E \to E$ in $(0,\,0)$, since that demands that for each neighbourhood $N$ of $0$ in $E$, there is a neighbourhood $D\times M$ of $(0,\,0)$ such that $D\cdot M \subset N$. $D$ contains a disk $D_\varepsilon = \{z \colon \lvert z\rvert < \varepsilon\}$ and $D_\varepsilon \cdot M$ is easily verified to be balanced.
Now, let's finish the proof.
If $\mathcal{T}$ is Hausdorff, so is $\mathcal{T}_V$ (trivial verification).
The closed unit ball $B$ in $\mathbb{K}^n$ is compact in $\mathcal{S}_n$, and so is its boundary $S$. Since $\Phi$ is continuous, and $\mathcal{T}_V$ Hausdorff, $\Phi(B)$ and $\Phi(S)$ are compact (in $\mathcal{T}_V$).
Since $0 \notin \Phi(S)$, for each $x \in \Phi(S)$, there is a balanced neighbourhood $U_x$ of $0$ and a neighbourhood $W_x$ of $x$ such that $U_x \cap W_x = \varnothing$.
Since $\Phi(S)$ is compact, there exist finitely many $x_1,\, \dotsc,\, x_k$ such that $\Phi(S) \subset \bigcup\limits_{i = 1}^k W_{x_i}$. Then $U := \bigcap\limits_{i=1}^k U_{x_i}$ is a balanced ($\mathcal{T}_V$) neighbourhood of $0$ that doesn't intersect $\Phi(S)$.
$\Phi^{-1}(U)$ is a balanced ($\mathcal{S}_n$) neighbourhood of $0$ that doesn't intersect $S$, hence $\Phi^{-1}(U) \subset \overset{\circ}{B}$, whence $U \subset \Phi(\overset{\circ}{B})$.
By linearity, it follows that for every $\mathcal{S}_n$-neighbourhood $N$ of $0$, $\Phi(N)$ is a $\mathcal{T}_V$-neighbourhood of $0$.
That in turn implies that $\Phi$ is open.
An open and continuous bijection is a homeomorphism, hence $\mathcal{T}_V = \mathcal{S}_n$. $\qquad$ c.q.f.d.
