Operator is continuous by boundedness? I have a fairly simple question, I just want to make sure I don't oversee something:

Let $l^\infty$ denote the vector space of all bounded sequences equipped with the $\sup$-norm and $c_0$ the subspace of all sequences converging to $0$. Define $$A:l^\infty \to c_0; \quad Ax:= (k^{-1}x_k)_{k \in \mathbb N}$$
  Clearly, $A$ is a well-defined linear operator. Show that $A$ is continuous.  

Now my solution is simply the observation that
$$\lVert Ax \rVert = \sup_k \lvert k^{-1}x_k \rvert \leq \lVert x \rVert_\infty$$
which means $A$ is bounded and, therefore, continuous. In the solution they make use of sequential continuity but essentially use the same inequality. Did I miss something?

Promoting comment to answer: your solution is fine.  This question is really easy.
(It would be a little bit harder if you didn't know the theorem that bounded operators are continuous, so you might want to double check that you're "allowed" to use that fact.)
