Show that the mapping $x^*\mapsto x^*(x)+r$ is weak$^*$ continuous Currently I am reading Behrend's $M$-Structure and Banach-Stone Theorem. The author introduced the following terminology at page $44,$ definition $2.13.$

Let $X$ be a Banach space, $x\in X, r\geq 0.$ Let 
  $$K(x,r) = \{(x^*,x^*(x)+r)\in X^*\times \mathbb{R}: x^*\in B_{X^*}\}.$$

Then the author mentioned that $K(x,r)$ is weak$^*$-compact due to the following: 

Since $B_{X^*}$ is weak$^*$-compact and the mapping $x^*\mapsto x^*(x)+r$ is weak$^*$ continuous, therefore $K(x,r)$ is compact. 

My question is, 

Question: How to show that the mapping $x^*\mapsto x^*(x)+r$ is weak$^*$ continuous?

To show the above, I think we need to show that if the net $\{x^*_\alpha\}_\alpha$ converges to $x^*$ in weak$^*$-topology, then $x_\alpha^*(x)+r$ converges to $x^*(x)+r$ in norm.
So by assumption, for any $x\in X,$ we have 
$$x^*_\alpha(x)\overset{|\cdot|}\to x^*(x).$$
Then 
$$|(x_\alpha^*(x)+r)-(x^*(x)+r)| = |x^*_\alpha(x)-x^*(x)|\to 0. $$
Therefore, the map is weak$^*$ continuous. 
Is my reasoning correct?
EDITED: After proving that the mapping is weak$^*$ continuous, how to see that the set $K(x,r)$ is compact? Is it because the set $K(x,r)$  can be written as Cartesian product between domain and the range of the weak$^*$ continuous map, and since both domain and range are compact in their weak$^*$ topology, therefore by Tychonoff's theorem, the set $K(x,r)$ is compact.

Yes, that's correct. "Weak$^*$-convergence" is a fancy name that makes sense for several reasons; but it is important to always remember that it is just pointwise convergence. 
The set $K(x,r)$ is compact, being a continuous image of a compact set.
