More impressively, you can have function that is continuous, but you cannot find a connected path on it (i.e. not path connected). In plain words, if anyone told you "a function is continuous when you can draw it without lifting your pen". They have lied to you.
You are misrepresenting an analogy as a lie. Besides that, in the context where the claim is typically made, the analogy is still pretty reasonable and your example is just plain wrong.
People are talking about continuous maps on subsets of R into R with this analogy basically always (i.e., during a typical calc 1 or precalc class). The only real issue are domain requirements in such a context. You need connectedness in the domain or else you're just always forced into lifting your pen.
There are a couple other requirements you could add as well. You might also want to avoid unbounded domains since you can't physically draw an infinitely long curve. Likewise you might want to avoid open endpoints or else things like 1/x on (0,1] become a similar kind of problem. But this is all trivial to avoid by saying "on a closed and bounded interval" and the analogy is still fairly reasonable without them so long as you keep the connectedness requirement.
For why your example is just wrong in such a context, say we're only dealing with continuous maps on a connected subset of R into R. Recall the connected sets in R are just intervals. Recall the graph of a function f with domain X is the set {(x,f(x)) : x is in X}. Do you see why the graph of such a function is always path connected? Hint: Pick any pair of points on this graph. Do you see what path connects those two points?
Once you want to talk about continuous maps between more general topological spaces, things become more complicated. But that is not within the context in which this analogy is made.