Prove that $\lim_{n\rightarrow \infty} T_n(x) = T(x) \iff \lim_{n\rightarrow \infty}\sup_{x \in K}\|T_n(x) - T(x)\| = 0$ Suppose $X$ and $Y$ are Banach spaces and $T:X \rightarrow Y$ is a BLO and $K$ is a compact subset of $X$. Prove that:
$$\lim_{n\rightarrow \infty} T_n(x) = T(x) \iff \lim_{n\rightarrow \infty} \sup_{x \in K}\|T_n(x) - T(x)\| = 0$$
the '$\Leftarrow$' implication is trivial, but how do I even start with the $\Rightarrow$ implication?
A few hints I got:
We know $\sup_{n\geq1}\|T_n\| < \infty$ and for any $K$: there is a collection of points $\{x_1, \dots, x_l\}$. Such that:
$$\sup_{x \in K} \inf_{1\leq i \leq l} \|x-x_i\| < \epsilon$$
 A: As @AlexanderFrei pointed out, it should read
$$\forall x: \lim_{n \to \infty} T_n(x) = T(x) \iff \forall K \subseteq X \, \text{compact}: \lim_{n \to \infty} \sup_{x \in K} \|T_n(x)-T(x)\| = 0.$$
The implication "$\Rightarrow$" is trivial, just choose $K= \{x\}$ for fixed $x \in X$. It remains to prove "$\Leftarrow$".
Suppose that $T_n(x) \to T(x)$ for each $x \in X$ and let $K$ be a compact set. Then for any $\epsilon>0$ the family of open sets
$$B(x,\epsilon):= \{x' \in X; |x-x'| <\epsilon\}, \qquad x \in K,$$
is a covering of $K$, and therefore there exist $x_1,\ldots,x_k \in K$ such that
$$K \subseteq \bigcup_{i=1}^k B(x_i,\epsilon). \tag{1}$$
Since we know that $T_n(x_i) \to T_n(x_i)$ for all $i=,1\ldots,k$, we can choose $N \in \mathbb{N}$ such that
$$\|T(x_i)-T_n(x_i)\| \leq \epsilon \quad \text{for all $n \geq N$ and $i \in \{1,\ldots,k\}$.} \tag{2}$$
Now let $x \in K$. By $(1)$, there exists $i \in \{1,\ldots,k\}$ such that $x \in B(x_i,\epsilon)$. By the triangle inequality,
$$\|T_n(x)-T(x)\| \leq \|T_n(x)-T_n(x_i)\|+\|T_n(x_i)-T(x_i)\|+\|T(x_i)-T(x)\|.$$
Now use $(2)$, $\|x-x_i\| < \epsilon$ and the boundedness of the operator to conclude
$$\|T_n(x)-T(x)\| \leq \left( \sup_n \|T_n\|+\|T\| \right) \epsilon + \epsilon$$
for all $n \geq N$. Since $N$ does not depend on $x$, this proves $$\sup_{x \in K} \|T_n(x)-T(x)\| \xrightarrow[]{n \to \infty} 0.$$
