For the true part(s), you may want to be more precise and refer to specific results. I can't fill in the details for you without knowing more about what class you're in (calculus 1? real analysis? topology?).
Your part (a) is wrong because $f$ is not assumed to be continuous. You can cook up as nasty a function as you want as a counterexample.
For parts (c), (d), and (e), you should probably write down counterexamples. For instance, in (c), you note: "$f$ is not on a closed interval." This is begging the question, because that's more or less what you're being asked to justify. There are many continuous functions on an open interval that are bounded, so you should explicitly produce a function on an open interval that is not bounded.
I think you are greatly over complicating the statement; take a step back.
A complex number can be written as $z=x+iy$ and a complex function with complex output is given by $f(z)=u(x,y)+iv(x,y),$ where $u:\mathbb{R}^2\to\mathbb{R}$, $v:\mathbb{R}^2\to\mathbb{R}$. Note that if $v(x,y)=0$ for all $(x,y),$ then $f$ is just real-valued.
The magnitude (or modulus) is given by $|f(z)|=\sqrt{[u(x,y)]^2 + [v(x,y)]^2},$ which is a real number. Show that the function $(x,y)\mapsto\sqrt{[u(x,y)]^2 + [v(x,y)]^2}$ is continuous, which is true since $f$ is continuous.
If all $(x,y)\in K,$ where $K$ is a compact subset of $\mathbb{R}^2$, then you can just apply the extreme value theorem from $\mathbb{R}^2\to\mathbb{R}.$ Somewhere in $K$, $|f(z)|$ has maximum modulus.
Best Answer
I guess I found the counterexample myself. Consider the following function defined on $[0,1]$. $$f(x) = \left \{ \begin{array}{ll} 2x & \mbox{ if } 0 \leq x < 0.5 \\ 0.5 & \mbox{ if } 0.5 \leq x \leq 1 \end{array} \right . $$
The sup of the function is $1$, but there is no point in the domain $[0,1]$ that 'attains' this point.