I am working on dynamical systems (more specifically Sharkovskii) and I have to show there exists a $3$-cycle for a continuous function with $f(a) = b, f(b) = c, f(c)= d, f(d) = e, f(e) = a$ where $a<b<c<d<e$.
Now I wonder if my approach works. My idea is, since $f$ is continuous, that we know intervals map to the next interval (except for the last one), and thus we can make use of subsections of each interval. Thus I do the following:
$\exists B \subseteq [b,c]$ with $ f(B) = [c,d]$, also $\exists C \subseteq [c,d]$ with $ f(C) = [d,e]$ and since $f[d,e] = [a,d]$, there also exists some $D \subseteq [d,e]$ with $f(D) = [b,c]$.
Hence there is some subset of $D$, which we name $E$, such that $$f^{3}(E) = f^{2}[b,c] = f[c,d] = [d,e] \supseteq E,$$
which yields we have a $3$-cycle.
Now since my experience is limited, I wonder if the above holds and I would really appreciate some feedback.
Your last chain is wrong in this way. You only get inclusions, as the images of intervals may be larger than the intermediate value theorem requires.
So you get $$ f^3(D)=f^2([b,c])\supseteq f^2(B)=f([c,d])\supseteq f(C)=[d,e]\supseteq D $$ This means that $(f^3|_D)^{-1}$ maps $D$ into itself. Now make sure that $f^3$ is monotonic on $D$ then the inverse is continuous and thus has a fixed point, giving the 3-cycle. Or argue that the nested sequence of pre-images $(f^3|_D)^{-n}(D)$ has a non-empty limit.