Let $A$ be a C*-algebra and $\phi: A_+ \to [0,\infty]$ be a weight on $A$ (see for example Pedersen's book $C^*$-algebras and their automorphism groups, section 5).
Assume the following : If $A_+ \ni x_i \uparrow x$ then $\phi(x_i) \uparrow \phi(x)$.
Then the following seems to be true:
Claim: $\phi^{-1}([0,\alpha])$ is closed, for any $\alpha \in [0,\infty)$, i.e. $\phi$ is lower semi-continuous.
If we now take a say sequence $(x_n)$ in $A_+$ with $x_n \to x$ and $\phi(x_n) \leq \alpha$ we want to show that $\phi(x) \leq \alpha$.
The problem is then to find an increaing sequence $(y_n)$ in $A_+$ which still converges to $x$ and such that $\phi(y_n) \leq \alpha$ for all $n$.
I don't think there is a direct argument (of course I could be wrong about it).
The way I know is as follows:
A normal weight is an infinite sum of positive normal functionals. So, via an approximation argument, the question is reduced to functionals.
For a positive functional, normality is equivalent to wot/sot continuity on bounded sets (note that you are allowed to take sequences, by the wot-compactness of the unit ball). This is done for instance in Theorem 7.1.12 in Kadison-Ringrose.
Edit: a proof, for a weight $\phi$, of the equivalence between
$\phi$ is normal
$\phi$ is wot-lower-semicontinuous
was given by Uffe Haagerup, J. Functional Analysis 19 (1975), 302–317. The proof also takes most of Chapter 1, Section 1, in Stratila's Modular Theory in Operator Algebras.