Let me starts with some definitions : $f\colon \mathbb R\to \mathbb R $ is everywhere surjective if $f[I]=\mathbb R$ for every nonempty open interval $I.$ $f$ is Darboux function if $f$ satisfies the intermediate value property. The cardinality of the family of all everywhere surjective from $\mathbb R$ to $\mathbb R$ is $2^{c}.$ ($c$ is continuum). Also , notice that everywhere surjective is equivalent for saying $f^{-1}(r)$ is dense for all $r\in\mathbb R.$ My question is:
Show that if $h$ is continuous function with the property that every nonempty open interval contains a nonempty subinterval on which $h$ is constant then $f+h$ is everywhere surjective for every everywhere surjective function $f.$
I have done a lot of construction functions like this by using transfinite induction but in this one I could not since the cardinality of everywhere surjective is $2^c$ and the cardinal of $\mathbb R$ is $c.$ I think, the transfinite induction would not work for this problem. I am trying to do it like that, BWOC, fix $f$ is everywhere surjective and assume $f+h$ is not everhwere surjective, that is, $(f+h)^{-1}(r)$ is not dense from some $r\in\mathbb R.$ Then there exists an open interval $(a,b)$ such that $ (f+h)^{-1}(r)\cap (a,b)=\emptyset$, that is, for all $x\in(a,b)$ $$f(x)+h(x)\neq r$$ $$ h(x)\neq r-f(x)$$ since $(c,d)\subset(a,b)$ such that $h(x)=y$ for all $x\in(c,d)$. So, $$f(x)\neq r-y$$ for all $x\in(c,d)$ which is impassable since $f(c,d)=\mathbb R.$ I checked my argument is correct
