$T$ is continuous if and only if $\ker T$ is closed

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.

I am doing for $Y=\mathbb{R}$; Clearly if $f$ is continuous then its kernel is closed set. for the converse, assume that $f\neq0$ and that $f^{-1}(\{0\})$ is a closed set. Pick some $e$ in $X$ with $f(e)=1$. Suppose by way of contradiction $||f||=\infty$. Then there exists a sequence $\{x_n\}$ in $X$ with $||x_n||=1$ and $f(x_n)\ge n$ for all $n$. Note that the sequence $\{y_n\}$ defined by $y_n=e-\frac{x_n}{f(x_n)}$, satisfies $y_n\in f^{-1}(\{0\})$ for all $n$ and $y_n\rightarrow e$. Since the set $f^{-1}(\{0\})$ is closed it follows that $e$ must belong to it and consequently $f(e)=0$ which is a contradiction. Thus $f$ is a continuous linear functional.