Closed kernel implies linear operator to finite-dimensional space is continuous First I'd like to say that although I have seen similar questions here, all answers I could find used the quotient space, which we have not seen in class. Apologies if it has already been answered without using quotient spaces and I didn't find it. The result I want to prove is the following

Let $E$ and $F$ be normed spaces, with $F$ finite-dimensional, and let $T: E \to F$ be a linear operator. Show that if $\mathrm{ker}(T)$ is closed then $T$ is continuous.

We have already proved it holds for a functional $\phi: E\to \mathbb{R}$ or  $\mathbb{C}$ so I thought maybe I could extend that result but couldn't quite see how.
 A: Let me assume that $T$ is surjective (if $T:E\to R(T)$ is continuous then it is continuous from $E$ to the larger space $F$). Let me assume that $R(T)=\mathbb F^k$ for some $k\ge0$. The general case can be proven using that $F$ is isomorphic to $\mathbb F^k$ (with $\mathbb K=\mathbb R$ or $\mathbb C$).
Assume $\ker T$ is closed and $T$ is unbounded/not continuous. Then there is a sequence $x_n \to 0$ such that $\|Tx_n\|=1$. Since the image space is finite-dimensional we can assume wlog $Tx_n \to y$. Since $T$ is surjective, for each unit vector $e_i$ there is $f_i$ such that $Tf_i = e_i$.
Because $T(x_n) = \sum_{i=1}^k T(x_n)^te_i \cdot e_i$, it follows
$$
x_n - \sum_{i=1}^k T(x_n)^te_i \cdot f_i \in \ker T.
$$
Since $x_n\to 0$, $Tx_n \to y$, and $\ker T$ is closed it follows
$$
 - \sum_{i=1}^k y^te_i \cdot f_i \in \ker T.
$$
Hence
$$
0 = T( \cdot f_i) = \sum_{i=1}^k y^te_i \cdot T(f_i) = \sum_{i=1}^k y^te_i \cdot e_i= y,
$$
which is a contradiction to $\|y\|= \lim \|Tx_n\|=1$. And $T$ is bounded.

I tried to mimic the proof of the $k=1$ case without using arguments similar to the quotient space proof. It was surprisingly hard to came up with a proof :)
