Give an example of $f$ such that $\lim_{x \to 0} f(x) = 0$ but $\lim_{x \to 0} (f \circ f)(x) \neq 0$ Unfortunately I don't have much to offer with regards to 'work in progress'. I've tried a dozen drawings and know I have find $f$ which satisfies:
$\begin{equation}
\begin{split}
&\forall \epsilon > 0 \ \exists \delta > 0 \ \forall x: 0 < |x| < \delta \implies |f(x)| < \epsilon \ \ \text{and} \\
&\exists \epsilon' > 0 \ \forall \delta>0 \ \exists x: (0 < |x| < \delta \implies |f(x)| < \epsilon) \ \ \text{but} |f(f(x)| \geq \epsilon' 
\end{split}
\end{equation}$
 A: Nevermind:
$f(x)=\begin{cases}
0 &x\neq 0\\
1 &x=0
\end{cases}$
The important bit is $0<|x|$ instead of $0\leq|x|$. The short notation $\lim_{x\to 0}$ is a bit misleading here. Look at the actual definition.
$$\forall \varepsilon>0 \exists\delta>0 \forall x:0<|x|<\delta⟹|f(x)|<\varepsilon$$
Now let $\varepsilon>0$ be arbitrary, then for $\delta=1$, I can select any $0<|x|<\delta$ and have:
$$|f(x)-0|=|f(x)|=0$$
Since we are explicitly not allowed to select $x=0$

The exercise is actually quite clever. It teaches you the difference between a removable discontinuity (cf. removable singularity) and a continuous function. 

Original:
Such an f does not exist. Assume such an $f$ exists. Then for any sequence $x_n\to 0$, we know, that $f(x_n)\to0$. But if the second condition would hold, then for any $\delta>0$ I can find an $x$ with $|x|<\delta$ with $|f(f(x))|>\varepsilon'$. So I can find a sequence $x_n$ with 
$$|x_n|<1/n \text{ and }|f(f(x_n))|>\varepsilon'.$$
Well now define $y_n=f(x_n)$. Then we know that $y_n\to 0$ due to condition one. But this implies (also due to condition one), that $f(y_n)\to 0$. Which is a contradiction.
Essentially the first condition says that $f$ is continuous in $0$. And thus $f\circ f$ is also continuous in $0$, since $f(0)=0$
A: There is no such $f$ with $f(x)\ne0$ for all $x$.
Assume $\lim_{x\to0}f(x)=0$, and $f(x)\ne0$ for all $x$. Let an $\epsilon>0$ be given. Then there is a $\delta>0$ such that $0<|x|<\delta$ implies $0<|f(x)|<\epsilon$. Furthermore there is a $\delta'$ such that $0<|x|<\delta'$ implies $0<|f(x)|<\delta$. This allows to conclude that
$$0<|x|<\delta'\quad\Rightarrow\quad 0<|f(x)|<\delta\quad\Rightarrow \quad0<\bigl|f(f(x))\bigr|<\epsilon\ .$$
When $f$ may take the value $0$ things are different. Consider
$$f(x):=x\sin{1\over x}\quad(x\ne0),\qquad f(0)=1\ .$$
Then $\lim_{x\to0}f(x)=0$, but there is a sequence $x_n:={1\over n\pi}\to0$ with $\lim_{n\to\infty}f\bigl(f(x_n)\bigr)=1$, and similarly a sequence $y_n\to0$ with $\lim_{n\to\infty}f\bigl(f(y_n)\bigr)=0$. Therefore the $\lim_{x\to0}f\bigl(f(x)\bigr)$ does not exist.
