I learned what a Veblen function is a few days ago. It is a function from ordinals to ordinals that's strictly monotonic increasing and continuous.
The Wikipedia article is a little bit difficult to understand, mostly because I don't really know what the relevant notion of continuity is when discussing ordinal-valued functions.
So, are $\alpha \mapsto \beth_\alpha$ and $\alpha \mapsto \aleph_\alpha$ Veblen functions? They are monotonic increasing in the sense that don't reverse $\le$ and are injective, but I don't know whether they're continuous or not. My gut says that these functions spread out the domain too much and their images, intuitively, have tons of gaps, but I don't know how this relates to the relevant notion of continuity.
First of all, a terminological quibble. I've never heard the term "Veblen function" used to refer to any strictly increasing continuous function. Rather, in my experience it's reserved for those which arise via "climbing up" some system of ordinal notations.
That said, your question still makes perfect sense: you're asking whether the $\aleph$- and $\beth$-functions are continuous. The answer is yes, and this is directly from how they're defined at limit ordinals: both $\aleph_-$ and $\beth_-$ are examples of functions $F$ satisfying the limit rule $$F(\lambda)=\sup\{F(\alpha):\alpha<\lambda\}$$ for each limit ordinal $\lambda$. Any nondecreasing function satisfying this limit rule is continuous (it's a good exercise to show that I do need "nondecreasing" here - if I want to generalize to all functions I need a more complicated limit rule).
Keep in mind that in the ordinals, every successor ordinal is isolated, so interesting topological behavior is really focused on limit ordinals. The fact that e.g. $\beth_\alpha$ is often much bigger than $\alpha$ doesn't matter - all that matters is that $\beth_\lambda$ doesn't "jump past" the sequence $(\beth_\alpha)_{\alpha<\lambda}$ when $\lambda$ is a limit.