This is a question on a proof in the paper "Baire Category, Probabilistic Constructions and Convolution Squares" by TW Körner (Theorem 4.10 page 13). The theorem states that in the complete metric space $\mathcal{G}= \{ (E_1,E_2): E_1,E_2 \text{ are compact subsets of} \ \mathbb{T}, E_1+E_2 =\mathbb{T} \}$ with metric $d((E_1,E_2),(F_1,F_2))= \Delta(E_1,F_1) + \Delta(E_2,F_2)$ ($\Delta$ denotes the Hausdorff distance), the set of elements $(E_1,E_2)$ that are pairs of Kronecker sets is dense and open in $\mathcal{G}$.
Here $S(\mathbb{T})=\{ f \in C(\mathbb{T}): |f|=1 \}$ and $\{ f_j \} $ a countable dense subset of $S(\mathbb{T})$.
The statement I don't understand is this:
"By the uniform continuity of $f_j$ and the intermediate value theorem, any sufficiently large M will have the property that the equation $f_j(t)=e^{2\pi i M t}$ will have at least one solution in any closed interval $I$ of length $\frac{\epsilon}{ 4}$."
If one chooses an irrational number $\xi \in (0,1)$ the sequence $\{ e^{2\pi i n \xi} \}_{n\geq 1} $ will be dense in $\mathbb{T}$ by Weyl's equidistribution theorem. Does it have something to do with this?
The intuition is quite simple: By selecting $M$ really large, over any interval of width $\frac{\epsilon}{4}$ the map $t \mapsto e^{2\pi iMt}$ will complete as many revolutions of the circle as you like; but on the other hand, there are only so many complete revolutions of the circle (if any) that a given continuous function $f_j$ can complete over an interval of width $\frac{\epsilon}{4}$. Hence, by selecting $M$ large enough, you can guarantee that on any interval of width $\frac{\epsilon}{4}$ the two functions will hit each other (in fact, as many times as you like).
But now let's make this more rigorous:
For any $t \in \mathbb{R}$, write $[t] \in \mathbb{T} \simeq \mathbb{R}/\mathbb{Z}\ $ for the projection of $t$.
For all $t$, let $\, D([t]):=F(t+\frac{\epsilon}{4})-F(t)$,$\,$ where $F(\cdot)$ is a continuous function satisfying $\,f_j([t])=e^{2\pi iF(t)}$.$\,$ Note that $D(\cdot)$ is a continuous function on $\mathbb{T}$, and so since $\mathbb{T}$ is compact, $D(\cdot)$ is bounded.
So let $K<\infty$ be such that $D(\cdot)<K$ on the whole of $\mathbb{T}$.$\,$ Take $M$ sufficiently large that $M.\!\frac{\epsilon}{4}>K+1$.
Now let $\,I \subset \mathbb{T}\;$ be any arc of length $\frac{\epsilon}{4}$, and let $u \in \mathbb{R}$ be such that $[u]$ and $[u+\frac{\epsilon}{4}]$ are respectively the left and right endpoints of $I$. Fixing a function $F(\cdot)$ as above, let $n \in \mathbb{Z}$ be such that $Mu-n \in [F(u)-1,F(u))$. Then $$ M(u+\tfrac{\epsilon}{4})-n \,\geq\, F(u)-1+M.\!\tfrac{\epsilon}{4} \,>\, F(u)+K \,>\, F(u)+D([u]) \,=\, F(u+\tfrac{\epsilon}{4}). $$ So then, $\ Mu-n<F(u)\ $ while $\ M(u+\frac{\epsilon}{4})-n>F(u+\frac{\epsilon}{4})$.$\,$ Hence, by the "intermediate value theorem for two functions" (as linked in your comment), there exists $t \in (u,u+\frac{\epsilon}{4})$ such that $Mt - n = F(t)$ and therefore $f_j([t])=e^{2\pi i Mt}$.$\,$ Since $t \in (u,u+\frac{\epsilon}{4})$, we have that $[t] \in I$.