Understanding the proof of $l_2$ being complete.

Question $1)$ you have a sequence of functions $f_n\in l_2$ and you prove that it convergence (pointwise) to a function $f$, then you prove that this convergence is also an $l_2$ convergence. So to be completely formal you have to show that $f$ is also in $l_2$. (it is almost like that $1$ is a limit of $1-1/n$ but $(0,1)$ is not complete because $1$ is not in it).

Question $2)$ what's an element in $l_2$? there are two (equivalent) ways to answer that. The first, is that elements in $l_2$ are sequences $a_n$ such that $\sum_{n=1}^\infty |a_n|^2<\infty$. The second is that elements in $l_2$ are functions $f:\mathbb{N}\rightarrow\mathbb{C}$ satisfying that $\sum_{n=1}^\infty |f(n)|^2<\infty$. The notation $f(k)$ is clear when using the second version, when using the first version the notation $f(k)$ just means that it is the $k$'th coordinate of the sequence.

Question $3)$ this is the tricky part because we have two different norms in here.

The first is the $l_2$ norm that is $\|f\|_2 =\sum_{n=1}^\infty |f(n)|^2$ and the second is the norm on $\mathbb{C}$, for every given $k$, $f(k)$ is an element in $\mathbb{C}$. The inequality $|f_n(k)-f_m(k)|\leq \|f_n-f_m\|$ is in fact infinitely many inequalities once for every $k$. Remember that the sequence $f_n$ was taken to be a Cauchy sequence in $l_2$, hence the inequality implies that for every given $k$ the sequence $f_n(k)$ ($k$ is given, $n$ runs to infinity) is a Cauchy sequence (now in $\mathbb{C}$), therefore you can use the fact that $\mathbb{C}$ is complete.

Let's start with this $k$ in $f_n(k)$: The sequence $f_n\in l_2$ is actually a sequence of sequences, so every $f_n$ is a sequence, for example $$(f_1)_{k\in\mathbb{N}}=((f_1)_1,(f_1)_2,(f_1)_3,\ldots).$$ To simplify the expression and have a connection to maps $\mathbb{N}\rightarrow\mathbb{R}$ we write $f_1(k)$ instead of $(f_1)_k$.

Next point: Why step 3 is necessary: What we have so far, is that $f_n(k)\rightarrow f(k)$ for every $k$. This is called pointwise convergence. Please note, that this convergence is not uniform in $k$. In detail this can be explained by the definition of pointwise convergence: $$\forall \varepsilon>0\ \forall k,\ \exists n_0=n_0(\varepsilon,k)\in\mathbb{N} \mbox{ such that } \forall n\geq n_0:\ |f_n(k)-f(k)|<\varepsilon.$$ Please note, that this $n_0$ is dependent on $k$. This is not the convergence we want. We want convergence in $l_2$, which is defined by convergence in the following norm $$\|(f)_{k\in\mathbb{N}}\|^2=\sum_{k=1}^\infty (f(k))^2.$$ So $f_n\rightarrow f$ in $l_2$ is by definition $$\forall \varepsilon>0\ \exists n_0=n_0(\varepsilon)\mbox{ such that }\forall n\geq n_0\ \|f_n-f\|_{l_2}^2:=\sum^\infty_{k=1}(f_n(k)-f(k))^2\leq \varepsilon.$$ Here we sum over all $k$ and $n_0$ is not dependent on any $k$. This is what is proven in step 3.

Your last point is now hopefully a bit clearer after you see the definition of the $l_2$ norm and yanko's answer.