The questions were asked almost 4 years ago and don't know if there is any interest - had to say something about Q2 since I don't see anything redeemable about it. If we take the statement of the question at face value, the conclusion is wrong if the continuous function is a constant, $f(x)=k, k\in\mathbb{R}$. Choose $\varepsilon < k$, then $X_1=[0,1]$; choose $\varepsilon\geqslant k$, then $X_1=\varnothing$. Suppose $[0,\infty)$ is a co-domain spec; one would hope so since a continuous mapping with compact support can not generate an unbounded range; its range is closed and bounded. So say, $f:[0,1]\to[a,b]$ is continuous and onto with $b>a\geqslant0$. If we now suppose $f$ is one-to-one, the conclusion fails once again. The possible sets are $X_1=\varnothing$, $X_1=[0,1]$, $X_1=[0,p)$ or $X_1=(p,1]$ where $p\in(0,1)$ and $f(p)=\varepsilon$. The conclusion is true for some continuous functions that are not one-to-one. For example, $f(x)=1-x^2$ on $[0,1]$ fits the bill; choose $\varepsilon\in(0,1)$ and $X_1=(-\sqrt{1-\varepsilon},\sqrt{1-\varepsilon})$. The statement, in general, is true for those continuous functions that are not one-to-one, which attain exactly one maximum on $(0,1)$ or if they attain more than one maxima on $(0,1)$, they attain a specific value in $(a,b)$ an even number of times.
Please point out any errors.