TT is continuous if and only if kerT\ker T is closed

Let $X,Y$ be normed linear spaces. Let $T: X\to Y$ be linear. If $X$ is finite dimensional, show
that $T$ is continuous. If $Y$ is finite dimensional, show that $T$ is continuous if and only if $\ker T$ is closed.

I am able to show that $X$, finite dimensional $\implies$ $T$ is bounded, hence continuous.

For the second part: This is what I have:

Suppose $T$ is continuous. By definition $\ker T = \{ x\in X : Tx = 0 \}$ , and so $\ker T$ is the continuous inverse of a closed set. Hence $\ker T$ is closed.

First, is what I have attempted okay. How about the other direction?

If $\ker(T)$ is closed then $X/\ker(T)$ is a normed vector space. Observe that the map $\overline{T}:X/\ker(T)\to Y$ given by $\overline{T}(x+\ker(T))=T(x)$ is a well-defined linear map by the first part $\overline T$ is continuous since $X/\ker(T)$ is a finite dimensional vector space (as it is isomorphic to a subspace of $Y$). Let $\pi: X \to X/\ker(T)$ denote the quotient map. Note that $T=\overline{T}\circ \pi$ hence $T$ is continuous since it is a composition of continuous functions.