Convergence of $A_nT$ to $AT$ in operator norm for compact $T$

You need $T$ to be compact because otherwise, by taking $T=\Bbb1$, the statement would imply that strong convergence is equivalent to norm convergence, this is not true.

Suppose $\|AT-A_nT\|\not\to0$, this means that there exists a sequence of unit vectors $x_n$ so that $\|(AT-A_nT)x_n\|>\epsilon$ for some $\epsilon$ and for all $n$. Now $T$ is compact so the image of the unit ball under it is pre-compact. This means that $Tx_n$ has a convergent subsequence, so lets actually just assume $Tx_n$ to be convergent with limit $x$.

From strong convergence of the $A_n$ you see that $\|(A-A_n)x\|\to0$. Now lets combine our information: $$\|(AT-A_nT)x_n\|=\|(A-A_n)(Tx_n-x+x)\|≤\|(A-A_n)x\|+\|A_n-A\|\,\|T x_n-x\|.$$ Now $\|A_n-A\|≤(\|A_n\|+\|A\|)$ is bounded by some constant by assumption. Every other term on the right converges to zero. This is a contradiction to $\|(AT-A_nT)x_n\|>\epsilon$ for all $n$.

In order to prove that a sequence $(S_n)_n$ of operators converges to $S$ in operator norm, we have to prove that $$\lim_{n\to\infty}\sup_{x:\lVert x\rVert\leqslant 1}\left\lVert S_nx-Sx\right\rVert =0.$$ The supremum is taken on the closed unit ball $B$, which can be "big", hence the convergence may fail. Here, we have to prove that $$\lim_{n\to\infty}\sup_{y\in T(B)}\left\lVert A_ny-Ay\right\rVert =0.$$ This can be done by using precompactness of $T(B)$: for any fixed $\varepsilon$, there exists a finite set $F\subset Y$ such that for any $y\in Y$, there exists $y'\in F$ such that $\lVert y-y'\rVert\lt \varepsilon$.