Orbits existing for all time For part $c)$ I understand why the above argument implies that no solution can ever tend to infinity. However I don't understand why this implies that solutions exist for all time. Why if a solution tended to infinity in the $x-y$ plane would it mean that solutions do not exist for all time?
It is true that an unbounded solution of an ODE can exist for all positive times, for instance $$\begin{cases}x'=x\\y'=0\\x(0)=y(0)=1\end{cases}$$

Has maximal solution $g(t)=\begin{pmatrix}e^t\\1\end{pmatrix}$.

What you are supposed to prove is, roughly speaking, that if a solution ceases to exist in a finite time $t_e$, then it must diverge at $t_e$.

What you need to use is the following result, which sometimes goes under the name of "Compact-Escape Theorem":


  Let $F:\mathbb{R}\times\mathbb{R}^n\rightarrow\mathbb{R}^m$ a $C^1$ function. Let $y$ a maximal solution of the ODE
  $$\begin{cases} y'=F(t,y)\\ y(0)=y_0\end{cases}$$
  Then, the orbit $\Omega_{y_0}:=\{(t,s)\in\mathbb{R}\times\mathbb{R}^m\ s.t.\ t\geq0\wedge s=y(t)\}$ is not contained in any compact subset of $\mathbb{R}\times\mathbb{R}^m$.


In your case (c), since the image of $y$ is contained in a square, its domain cannot be limited.
Take a look at the likewise quadratic ODE $y'=y^2$ which transforms by separation to
$$
\frac1{y_0}-\frac1{y(t)}=t-t_0\iff y(t)=\frac{y_0}{1-y_0(t-t_0)}
$$
These solutions have poles in finite time, thus the solution for an IVP will not exist for all time.