When statement is false?

Question

$S \subseteq \mathbb{R}$, then the statement "there exist a continuous function $f : S\to S $ such that $f(x) \neq x$, for all $x\in S $" is false for $S=$

$\hspace{1em}\begin{array}{rl} a. & [2,3] \\ b. & (2,3] \\ c. & [-3,-2] \cup [2,3] \\ d. & (-\infty,\infty) \end{array}$

If we find where $f(x)= x$ exists and continuous among these 4 options, we can find the answer. But, it holds on all these $S$.

Answer 1

a. is false. Argue by contradiction and consider $f(x)-x$ combined with the intermediate value theorem.

Edit: Details added as requested.

Let $d(x) := f(x)-x$. If $f(2)=2$, then we are done, so assume otherwise.
Then $f(2)>2$ (because $f(2)\geq 2$!) and so $d(2)>0$.
Now if $f(3)=3$, then we are also done, so assume otherwise.
Then $f(3)<3$ (because $f(3)\leq 3$!) and so $d(3)<0$.
Since $f$ is continuous, so is $d$.
By the Intermediate Value Theorem, there exists $x$ such that $d(x)=0$.
Hence $f(x)=x$ and we are finally done.

Answer 2

Partial answer. Since max_zorn has answered a) I will answer the remaining. For c) take $f(x)=-x$ and for d) take $f(x)=x+1$. Hence c) and d) are true. So is b) ; take $f(x)=2+\frac {(x-2)^{2}} 2$. If $f(x)=x$ then $(x-2)^{2}= 2(x-2)$ which implies $x-2=2$ but $x-2<1$. We have to make sure that $f$ maps $(2,3]$ into itself. Clearly, $f(x)>2$. Also $f(x)\leq 2+\frac 1 2 <3$.