If $T: X \to Y$ is norm-norm continuous then it is weak-weak continuous

Solution 1:

It is of course true that norm-continuous linear maps are weakly continuous. This follows from the fact that the weak topology is the initial topology w.r.t. to all continuous linear functionals, i.e. $\sigma(Y,Y^*)$ is the coarsest topology on $Y$ such that all $f\in Y^*$ are continuous. Then, by abstract nonsense, a map $T:E \to (Y,\sigma(Y,Y^*))$ (where $E$ is an arbitrary topological space) is continuous if (and only if) all compositions $f\circ T$ are continuous. For $E=(X,\sigma(X,X^*))$ you have that continuity because $f\circ T$ is norm-continuous and hence $\sigma(X,X^*)$-continuous.

You should be very careful with your sequential proof (the weak topologies are not metrizable if the spaces are infinite dimensional). There are Banach spaces where the weakly convergent sequences are always norm-convergent, the space $\ell^1$ of absolutely summable sequences is the most prominent example of such Schur-spaces. This means that the identity $(\ell^1,\sigma(\ell^1,\ell^\infty)) \to (\ell^1,$norm-topology$)$ is sequentially continuous but NOT continuous.