$ X $ and $ Y $ are topological vector spaces, and $ f $ is a linear map from $ X $ onto $ Y $ with $ \dim Y $ finite. The conclusion is that $ f $ is open.
I don’t know because it doesn’t allow me to use the open mapping theorem. I tried to consider the quotient space while the continuity of $ f $ is not mentioned. So I have no confidence to prove $ \pi $ is open (the quotient map).
The textbook I'm using is Rudin's Functional Analysis. Much appreciation for your kindness and answer.
Since $f$ is onto and $dim Y = n < \infty$, choose a basis $\{y_1, \dots, y_n\} $ of $Y$ and preimages $x_i \in X$ with $f(x_i) = y_i$. Let $M = \operatorname{span}\{x_1, \dots, x_n\}$. Then $f|_M : M \to Y$ is a linear isomorphism.
In finite-dimensional Hausdorff TVS, any linear isomorphism is a homeomorphism (unique Euclidean topology).
Let $W$ be a neighborhood of $0$ in $X$. Let $B_M$ be a neighborhood of $0$ in $M$(subspace topology) with $B_M \subset W$. Since $f|_M$ is a homeomorphism, $f(B_M)$ is a neighborhood of $0$ in $Y$.
Thus $f(W) \supset f(B_M)$ is a neighborhood of $0$ in $Y$. So $f$ is open at $0$, hence open.
