Let $B(X,Y)$ be the family of all bounded maps from $X$ to $Y,$ normed linear maps. Then, $\Vert \cdot \Vert,$ defined by $\Vert T \Vert =\sup\limits_{\Vert x \Vert\leq 1}\Vert T x \Vert,$ for arbitrary $T\in B(X,Y), $ is a norm on $B(X,Y). $
we have for arbitrary $T\in B(X,Y), $ \begin{align} \Vert T \Vert =\sup\limits_{\Vert x \Vert\leq 1}\Vert T x \Vert= \sup\limits_{\Vert x \Vert = 1}\Vert T x \Vert=\sup\limits_{x\neq 0} \frac{\Vert T x \Vert}{\Vert x \Vert}.\end{align}
MY TRIAL
1. \begin{align} \Vert T \Vert =0&\iff \sup\limits_{\Vert x \Vert\leq 1}\Vert T x \Vert =0\iff \Vert T x \Vert =0,\;\;\forall \,x\in X, \,T\in B(X,Y)\\& \iff T x=0,\;\;\forall \,x\in X, \,T\in B(X,Y) \\& \iff T =0,\;\;\forall \, \,T\in B(X,Y)\end{align} 2. \begin{align} \Vert kT \Vert =& \sup\limits_{\Vert x \Vert\leq 1}\Vert k T x \Vert \\=& |k|\sup\limits_{\Vert x \Vert\leq 1}\Vert T x \Vert \\=& |k|\Vert T \Vert,\;\;\forall \,\,k\in K, \,T\in B(X,Y)\end{align}
3. \begin{align} \Vert T+S \Vert =& \sup\limits_{\Vert x \Vert\leq 1}\Vert T x + S x \Vert \\\leq & \sup\limits_{\Vert x \Vert\leq 1}\left(\Vert T x \Vert + \Vert S x \Vert \right) \\=&\sup\limits_{\Vert x \Vert\leq 1}\Vert T x \Vert +\sup\limits_{\Vert x \Vert\leq 1} \Vert S x \Vert \\=&\Vert T \Vert+\Vert S \Vert,\;\;\forall \,T,S\in B(X,Y)\end{align} Kindly help check if this is correct. If not, corrections and alternative proofs will be highly welcome.
Your proof is fine except that you have to fill in some details on why $\sup\{\|Tx\|:\|x\|\leq 1\}$ implies $Tx=0$ for all $x$. You get $Tx=0$ for $\|x\|\leq 1$ and then you have to argue that if $x \neq 0$ then $\frac x {\|x\|}$ has norm $1$ so $T\frac x {\|x\|}=0$; finally, linearity of $T$ gives $\frac {Tx} {\|x\|}=0$, so $Tx=0$.