any real number between $0$ and $1$ is expressible as an (infinite) sum of particular fractions

The key to this problem is the fact that the half-open intervals $({1\over k}, {1\over {k-1}}]$ for integer $k > 1$ partition the half-open interval $(0, 1]$.

Given $x \in (0, 1]$, put $x_0 = x$ and $k_0 = 2$ (so that $x_0 \le {1\over {k_0-1}}$). I claim that we can find a unique sequence of integers $k_n$ with $1 < k_1 \le k_2 \le \ldots$ such that the following equations and inequalities hold for some real numbers $x_n$ : $$ \begin{align*} x_0 = {1 \over k_1} + {x_1 \over k_1} \quad& 0 < x_1 \le {1 \over {k_1 -1}} \\ &\vdots\\ x_{n-1} = {1 \over k_n} + {x_n \over k_n} \quad& 0 < x_n \le {1 \over {k_n -1}}\\ &\vdots \end{align*} $$ To see this, at each stage take $k_n$ to be the unique integer such that ${1 \over {k_n}} < x_{n-1} \le {1 \over {k_n - 1}}$. Then $x_{n-1} = {1 \over {k_n}} + \epsilon_n$ where $0 < \epsilon_n \le {1 \over {k_n - 1}} - {1\over k_n} = {1 \over {k_n(k_n-1})}$. Now if we put $x_n = k_n \epsilon_n$, $k_n$ and $x_n$ will meet the requirements ($k_n \ge k_{n-1}$, because $x_{n-1} \le {1 \over {k_{n-1}-1}}$).

Clearly we now have:

$$ x= x_0 = {1\over k_1}+{1\over k_1k_2}+\cdots+{1\over k_1k_2\cdots k_n} + {x_n \over k_1k_2\cdots k_n} $$

whence, given our bounds on the $k_n$ and $x_n$, we have that the r.h.s of the following equation converges to the l.h.s.:

$$ x = {1\over k_1}+{1\over k_1k_2}+\cdots+{1\over k_1k_2\cdots k_{n}} + \ldots $$

That gives existence of the above series expansion for $x$. For uniqueness, note that if we apply our construction to ${1\over l_1}+{1\over l_1l_2}+\cdots+{1\over l_1l_2\cdots l_{n}} + \ldots$, where $1 < l_1 \le l_2 \ldots$, then we will find $k_1 = l_1, k_2 = l_2, \ldots$.