This answer is in response to final sentence, "I would be very grateful to anybody who would write a new answer with a few lines of explanation as to what Sierpinski is actually doing". In fact, it is easy to construct power series converging on the circle of convergence, but are unbounded. For example,
$$
f(z)=\sum_{n=1}^\infty\frac1{n^5(1+in^{-3}-z)}
$$
defines a function whose power series expansion has radius of convergence 1 and converges everywhere on the unit circle, but is unbounded in a neighbourhood of 1.
A method of constructing such functions is as an infinite sum
$$
f(z)=\sum_{n=1}^\infty f_n(z).
$$
Here, $f_n(z)$ are chosen to have a power series expansion converging everywhere on the closed unit ball. Let $f^{(r)}_n(z)$ denote the sum of the first $r$ terms in the power series expansion of $f_n$. We need to arrange it so that $f^{(r)}(z)\equiv\sum_nf_n^{(r)}(z)$ converges on the closed unit ball, and that $f(z)=\lim_{r\to\infty}f^{(r)}(z)$ holds. That is, we need to be able to commute the limit $r\to\infty$ with the summation over $n$. A sufficient condition to be able to do this is that $\sum_n\sup_r\lvert f^{(r)}_n(z)\rvert < \infty$, for all $\lvert z\rvert\le1$. That this allows us to commute the summation with the limit is just a special case of dominated convergence.
Next, to ensure that $f(z)$ is unbounded on the unit ball, we want to choose $f_n$ such that there exists $q_n$ in the closed unit ball with $f_n(q_n)$ large, and such that it does not get cancelled out in the summation, so that $f(q_n)$ is large and diverges as $n\to\infty$.
For example, choose positive reals $\delta_n,\epsilon_n$ tending to zero, and setting $a_n=1+i\epsilon_n$, and
$$
f_n(z)=\frac{\delta_n}{a_n-z}=\sum_{m=0}^\infty \delta_na_n^{-m-1}z^m.
$$
These are all well-defined as power series with radius of convergence greater than 1. Furthermore, the partial sums are
$$
f^{(r)}_n(z)=\delta_n\frac{1-(z/a_n)^r}{a_n-z},
$$
which are bounded by $2\delta_n/\lvert a_n-z\rvert$. As $a_n\to1$, this is bounded by a multiple of $\delta_n$ for each fixed $z\not=1$, so the dominated convergence condition is satisfied when $\sum_n\delta_n$ is finite. On the other hand, if $z=1$, then $\lvert a_n-z\rvert=\epsilon_n$, so the dominated convergence condition is satisfied everywhere whenever $\sum_n\delta_n/\epsilon_n$ is finite.
Next, $f_n(z)$ achieves its largest value on the unit ball at $q_n=a_n/\lvert a_n\rvert$, and its real part there is given by
$$
\Re f_n(q_n)=\frac{\delta_n}{\sqrt{1+\epsilon_n^2}(\sqrt{1+\epsilon_n^2}-1)}\ge\frac{2\delta_n}{\epsilon_n^2\sqrt{1+\epsilon_n^2}}.
$$
As $f_m(z)$ has positive real part for all $m$, this bound also holds for $f(q_n)$, and we get that $f$ is unbounded whenever $\delta_n/\epsilon_n^2\to\infty$. These conditions are satisfied by taking $\epsilon_n=n^{-3}$ and $\delta_n=n^{-5}$.
Alternatively, for an example closer to Sierpinski's, consider choosing a sequence $a_n\to1$ on the unit circle and positive reals $K_n$, and set
$$
f_n(z)=K_n2^{-n}\sum_{k=0}^{2^n-1}a_n^{2^n-1-k}z^k=2^{-n}K_n\frac{a_n^{2^n}-z^{2^n}}{a_n-z}.
$$
The partial sums of the power series expansion of $f_n(z)$ are bounded by $2^{1-n}K_n/\lvert a_n-z\rvert$, so the dominated convergence condition is satisfied for $z\not=1$ so long as $\sum_n2^{1-n}K_n$ is finite. Sierpinski chooses $a_n=(n^2-1+2ni)/(n^2+1)$ so that $a_n-1$ goes to zero at rate $1/n$. The dominated convergence condition is therefore satisfied whenever $\sum_n2^{-n}K_nn$ is finite.
Now, $f_n(z)$ is maximized at $z=a_n$ where $\lvert f_n(a_n)\rvert=K_n$. So,
$$
\lvert f(a_n)\rvert\ge K_n-\sum_{m\not=n}\frac{2^{1-m}K_m}{\lvert a_m-a_n\rvert}.
$$
As $a_m-a_n$ is bounded below by a multiple of $1/m^2$, the summation on the right is bounded whenever $\sum_m2^{-m}K_mm^2$ is finite, and $f(a_n)$ is unbounded if we also take $K_n$ going to infinity. Sierpinski takes $K_n=n^2$ here. Finally, in Sierpinski's example, he multiplies $f_n$ by $z^{2^n}$. This changes nothing, except to separate out the non-zero terms of the power series of $f_n(z)$, so that the power series of $f(z)$ can be written easily term by term.
