A holomorphic map between complex tori (proposition 1.3.2 in Diamond–Shurman) The following is from Diamond and Shurman's A First Course in Modular Forms book:

I had studied Munkres Topology a few years ago but for lifting I had to review the materials again, but I still don't see the connection to this theorem! 
Especially, could someone at easiest possible way explain why $f_{\lambda}$  becoming constant? And, how $m \Lambda \subset \Lambda'$ holds? 
(They are underlined blue in the image above)
 A: $\newcommand{\Q}{\Bbb Q}
\newcommand{\N}{\Bbb N}
\newcommand{\R}{\Bbb R}
\newcommand{\Z}{\Bbb Z}
\newcommand{\C}{\Bbb C}
\newcommand{\A}{\Bbb A}
\newcommand{\L}{\Lambda}
$
For your 2nd question, notice that we have a commutative diagram
$$\require{AMScd} \begin{CD}
\C @>{\widetilde{\phi}}>> \C \\ @V{\pi}VV @VV{\pi'}V\\
\C / \L @>>{\phi}> \C / \L'
\end{CD}$$
Thus if $z \in \L$, then
$$\pi'(\widetilde{\phi}(z)) = \phi(\pi(z)) = \phi(\pi(0))= \pi'(\widetilde{\phi}(0))$$
which implies 
$\widetilde{\phi}(z) - \widetilde{\phi}(0) = mz+b-b = mz \in \L'$.
Hence, we proved $m \L \subset \L'$.

For your 1st question, the map $f = f_{\lambda} : \C \to \C$ is continuous and its image is a discrete set, since it takes values in $\L'$ which is a lattice (hence discrete). 
Indeed, we have 
$$\pi'(f(z)) = \phi(\pi(z+\lambda)) - \phi(\pi(z)) = 0,$$
thus $f(z) \in \L'$ for every $z \in \C$.
Since $\C$ is a connected topological space, so is $f_{\lambda}(\C)$. Finally, a non-empty discrete connected topological space must be just a single point (see here).
Thus $f_{\lambda}$ is a constant map.

