Recall the theorem :
$T_n \in B(X,Y)$ where $X,\ Y$ are Banachs, is strongly convergent iff (a) $ \parallel T_n \parallel $ is bounded (b) $T_nx$ is Cauchy where $x$ is in total subset $M$.
$B$ implies bounded.
Subset $M$ is total if ${\rm span}\ M$ is dense in $X$
${\bf Proof}$ : By 267 page in Kreyszig's book, $\Rightarrow$-part is trivial. And for $\Leftarrow$-part, we can define $T$ : $$ T_nx\rightarrow Tx $$ for any $x\in X$.
But how can we show that $$ \parallel T_n - T\parallel \rightarrow 0. $$
You cannot show that, because it is false.
Strong convergence does not mean that $\Vert T_n -T \Vert \rightarrow 0$. It just means that $T_n x \rightarrow Tx$ for all $x \in X$.
For an example, consider $X = C([0,1])$ and $Y = \mathbb{R}$ and $$T_n f := n^{-1} \cdot \int_0^{1/n} f(x) dx.$$
It is easy to see that $T_n f \rightarrow f(0) =: T f$ for all $f \in C([0,1])$ and that $\Vert T_n \Vert \leq 1$ for all $n$.
But $\Vert T_n -T \Vert $ does not converge to zero, as for any $n \in \mathbb{N}$ there is $f_n \in C([0,1])$ with $f \geq 0$, $f(0) = 0$, $f|_{[1/2n, 1/n]} \equiv 2$, i.e. $T f_n = 0$, but $T_n f_n \geq 1$.