Let $X$ be a Banach space and $\phi: X \rightarrow C$ be a bounded linear functional. Then $\phi$ is weakly continuous is equivalent to $\phi$ is norm continuous, right? Why?
The weak topology is weaker than the norm topology. This implies that every weakly continuous map (needn't even be linear) is norm-continuous.
For the other implication: The weak topology is (more or less by definition) the weakest topology making all norm-continuous linear functionals continuous. Therefore, the norm-continuous functionals are weakly continuous.