Show that a compact operator is bounded 
Definition: A linear operator $T: V \to W$ is compact if and only if the image of the unit ball in $V$ is precompact (= every sequence
has a cauchy subsequence $\iff $ totally bounded).
Prove: Let $T: V \to W$ be a compact linear operator. Show that $T$ is
bounded.

My attempt:
Suppose $T$ is not bounded. Then,
$$\forall M > 0: \exists v_M \in V: \Vert T v_M \Vert > M \Vert v_M \Vert$$
I then tried to construct a sequence without cauchy sequence in $\{Tv \mid \Vert v \Vert \leq 1\}$
So, let $p > q$. Then, $$\left\Vert T \frac{v_p}{\Vert v_p \Vert} - T \frac{v_q}{\Vert v_q \Vert}\right\Vert \geq  \left|\frac{\Vert Tv_p \Vert}{\Vert v_p \Vert} -  \frac{\Vert Tv_q \Vert}{\Vert v_q \Vert}\right|$$
but was unable to conlude something because off the minus sign.
Any ideas?
EDIT: This is not a duplicate, as other posts use other definitions of compact operators.

If $T$ is unbounded, there is a sequence $(x_n)_{n\in\mathbb N}$ of elements of the unit ball in $V$ such that$$(\forall n\in\mathbb N):\bigl\|T(x_n)\bigr\|>n.$$Therefore, the sequence $\bigl(\|T(x_n)\|\bigr)_{n\in\mathbb N}$ is not a Cauchy sequence.
