Showing that $\ker T$ is closed if and only if $T$ is continuous.

It looks good, and is the typical argument, except for two things, one of which is minor:

In the first argument, after you say "let $x_n\in\text{Ker}\,T$", you also need to add "and $x_n\rightarrow x$". The previous sentence does not give that to you.

The more egregious error: In the second argument, you need to state that there is a bounded sequence $(x_n)$ such that $|T(x_n)|\rightarrow\infty$; that is, there is a sequence $(x_n)$ with $\Vert x_n\Vert \le 1$ (say) with $|T(x_n)|\rightarrow \infty$. (And you might want to justify this; though, it's almost automatic from the definition of boundedness.) Without the boundedness of the $x_n$, you would not be guaranteed that the sequence $(x_n')$ converges to $a$.

There is a somewhat quicker route for the forward implication: $\{0\}$ is closed in $\Bbb R$, so, since $T$ is continuous, $T^{-1}(\{0\})$ is closed in $X$.

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.