Suppose $\mathcal{X}$ and $\mathcal{Y}$ are normed vector spaces and $T\in L(\mathcal{X},\mathcal{Y})$. Let $N(T)=\{x\in \mathcal{X}: Tx=0\}$ be the kernel.
a) $N(T)$ is a closed subspace of $\mathcal{X}$
Here is my proof:
For a sequence $x_n\in N(T)$, suppose $x_n\to x\in \mathcal{X}$, we have $\Vert T(x_n)-T(x)\Vert\leq \Vert T\Vert\Vert x_n-x\Vert\to 0$. So $\Vert T(x)\Vert=\Vert T(x_n)-T(x)\Vert=0$ which implies $T(x)=0$. Thus, $x\in N(T)$ which gives that $N(T)$ is closed.
Is it right?
b) There is a unique $S\in L(\mathcal{X}/ N(T), \mathcal{Y})$ s.t. $T(x)=S(\pi(x))$ where $\pi: \mathcal{X}\to \mathcal{X}/N(T)$ is the projection. Moreover, $\Vert S\Vert=\Vert T\Vert$.
But HOW TO prove b)?
Your backslash should be a slash, it's just the quotient. You define $$\tag1 S(x+N(T))=Tx. $$ This is well-defined because if $x+N(T)=y+N(T)$ then $x-y\in N(T)$ and so $Tx=Ty$. The equation $S\circ\pi=T$ is precisely $(1)$.
Finally, given $\varepsilon>0$, there exists $x$ with $\|x\|=1$ and such that $\|Tx\|>\|T\|-\varepsilon$. As $\pi$ is a contraction, $\|x+N(T)\|\leq 1$. So $$ \|S(x+N(T))\|=\|Tx\|>\|T\|-\varepsilon\geq(\|T\|-\varepsilon)\,\|x+N(T)\|. $$ Thus $\|S\|\geq\|T\|-\varepsilon$. As $\varepsilon$ was arbitrary, $\|S\|\geq\|T\|$. On the other hand, for any $y\in N(T)$ $$ \|S(x+N(T))\|=\|Tx\|=\|T(x+y)\|\leq\|T\|\,\|x+y\|. $$ Taking infimum over $y$, we get $$\|S(x+N(T))\|\leq\|T\|\,\|x+N(T)\|,$$ and so $\|S\|\leq\|T\|$.