Solution:

In the first proof of the first part, we denote for a real number $x$ let $[x]$ be the largest integer not greater than $x$ (floor function), the number $\langle x\rangle:=x-[x] \in[0,1)$ is called the fractional part of $x$.
We choose an integer $n>\sqrt{2} \varepsilon^{-1}$. By the pigeonhole principle there exist distinct indices $i, j \in\{0,1, \ldots, n\}$ such that $\langle i \sqrt{6}\rangle$ and $\langle j \sqrt{6}\rangle$ lie in the same one of the $n$ intervals
$$\left[0, \frac{1}{n}\right),\left[\frac{1}{n}, \frac{2}{n}\right), \ldots,\left[\frac{n-1}{n}, 1\right)$$
Then there exists an integer $a$ such that $|a-(i \sqrt{6}-j \sqrt{6})|<1 / n$ holds. From this, by multiplying by $\sqrt{2}$, it follows for $b:=2(j-i)$ that the inequality $|a \sqrt{2}+b \sqrt{3}|<\sqrt{2} / n<\varepsilon$ holds, which with $c=0$ leads to the inequality in the problem statement.

The second proof gets by without the pigeonhole principle: Since $0<\sqrt{6}-2<1$, we find a positive integer $k$ such that $X=(\sqrt{6}-2)^{k} \in(0, \varepsilon / \sqrt{3})$. Evidently there exist integers $a_{0}, b_{0}$, different from 0, such that $X=a_{0} \sqrt{6}+b_{0}$. We now set $a=3 a_{0}, b=b_{0}, c=0$, then we have $|a \sqrt{2}+b \sqrt{3}+c \sqrt{5}|=\left|\sqrt{3}\left(a_{0} \sqrt{3} \sqrt{2}+b_{0}\right)\right|=\sqrt{3} X<\varepsilon$.

For the (more difficult) second part of the problem, we assume that $a, b$ and $c$ satisfy the inequality. We first compute that the following expression is an integer.
$$\begin{aligned} Z & :=(a \sqrt{2}+b \sqrt{3}+c \sqrt{5})(a \sqrt{2}-b \sqrt{3}-c \sqrt{5})(a \sqrt{2}+b \sqrt{3}-c \sqrt{5})(a \sqrt{2}-b \sqrt{3}+c \sqrt{5}) \\ & =\left(2 a^{2}-\left(3 b^{2}+2 \sqrt{15} b c+5 c^{2}\right)\right)\left(2 a^{2}-\left(3 b^{2}-2 \sqrt{15} b c+5 c^{2}\right)\right) \\ & =\left(2 a^{2}-3 b^{2}-5 c^{2}\right)^{2}-60 b^{2} c^{2} \end{aligned}$$
This number is different from 0, for otherwise 60 would be the square of a rational number, which, however, due to the unique prime factorization of positive integers, is not the case. Thus $|Z| \geq 1$ holds. On the other hand, $|Z|$ can be estimated from above as follows
$$\begin{aligned} |Z| & <|a \sqrt{2}+b \sqrt{3}+c \sqrt{5}| \cdot(|a| \sqrt{2}+|b| \sqrt{3}+|c| \sqrt{5})^{3} \\ & <\varepsilon\left((2+3+5)^{\frac{1}{2}} \cdot\left(a^{2}+b^{2}+c^{2}\right)^{\frac{1}{2}}\right)^{3} \\ & <\left(\varepsilon^{1 / 3} \sqrt{10} \cdot \sqrt{3} \cdot \max (|a|,|b|,|c|)\right)^{3} \end{aligned}$$
where in the second step we used the Cauchy-Schwarz inequality. After rearranging, the claim follows.