A Darboux function is a function that has the intermediate value property. That is a function $f$ such that
$$ \forall a,b \in \mathbb{R} : f[a,b] \supseteq [f(a),f(b)] \cup[f(b),f(a)] $$
We define the sum of two functions as such
$$ (f+g)(x) = f(x)+g(x)$$
Now the question is:
If $f$ is a Darboux function and $g$ is a continuous function, must $f+g$ be a Darboux function as well?
This is a very good question that leads into some fairly hard mathematics: the answer depends on what set theoretic assumptions you are prepared to make. If you assume the continuum hypothesis, there are what are called universally bad Darboux functions $f$ such that $f+g$ is not Darboux for any non-constant continuous $g$. See https://www.encyclopediaofmath.org/index.php/Darboux_property and the references it cites.
Here is a counterexample; no fancy set-theoretic axioms needed. A Cantor set is a set homeomorphic to the standard Cantor set.
Let $\{I_1,I_2,I_3,\dots\}$ be the set of all rational open intervals; let $I_n=(a_n,b_n)$ and let $c_n=\frac{a_n+b_n}2$.
Choose pairwise disjoint Cantor sets $A_1,B_1,A_2,B_2,A_3,B_3,\dots$ with $A_n\subseteq(a_n,c_n)$ and $B_n\subseteq(c_n,b_n)$.
Define $\alpha_n=\min\{c_n,-n\}$ and $\beta_n=\max\{c_n,n\}$.
Define $f:\mathbb R\to\mathbb R$ so that $f[A_n]=[c_n,\beta_n]$, $f[B_n]=[\alpha_n,c_n]$, and $f(x)=x+1$ if $x\notin\bigcup_n(A_n\cup B_n)$.
Since every interval of $\mathbb R$ contains infinitely many rational intervals $I_n$, and since $f[I_n]\supseteq f[A_n]\cup f[B_n]=[\alpha_n,\beta_n]\supseteq[-n,n]$, the function $f$ takes all real values in every interval, and is therefore a Darboux function.
Let $g(x)=-x$ and $h(x)=f(x)+g(x)=f(x)-x$.
If $x\in A_n$, then $x\lt c_n\le f(x)$, so $h(x)\gt0$; i.e., $\{x:h(x)\gt0\}\supseteq\bigcup_nA_n$ is dense in $\mathbb R$. Similarly, if $x\in B_n$, then $x\gt c_n\ge f(x)$, so $h(x)\lt0$; i.e., $\{x:h(x)\lt0\}\supseteq\bigcup_nB_n$ is dense in $\mathbb R$. However, $f(x)\ne x$ for all $x$, so $h(x)$ is never zero. This shows that $h$ is not a Darboux function.
P.S. A commenter says I need to justify the nontrivial fact then the sequence of Cantor sets $A_n, B_n$ can be constructed. Suppose that $A_1,B_1,\dots,A_{n-1},B_{n-1}$ have already been constructed. Then the set $E_n=A_1\cup B_1\cup\cdots\cup A_{n-1}\cup B_{n-1}$ is nowhere dense, because the Cantor sets $A_i,B_i$ are nowhere dense, and a finite union of nowhere dense sets is nowhere dense. In particular $E_n$ is not dense in $(a_n,c_n)$, so the set $(a_n,c_n)\setminus E_n$ contains an interval, and we can construct a Cantor set $A_n$ inside this interval. Likewise, we can construct a Cantor set $B_n$ inside $(c_n,b_n)\setminus E_n$.
