Let $H$ be a Hilbert space and $\{e_j\}_{j \in \Gamma} \subset H$ be a orthonormal system. Then $\{e_j\}_{j \in \Gamma}$ is called an orthonormal basis of $H$ if
$\overline{\operatorname{span}\{e_j\: | \: j \in \Gamma \}} = H$.
Furthermore, we have $\{e_j \: | \: j \in \Gamma \}^\perp = \overline{\operatorname{span}\{e_j\: | \: j \in \Gamma \}}^\perp$
I have a lot of trouble to understand what would go wrong if we would not consider the closure in these both statements. Is there a counterexample for this? I appreciate any help.
That is a very good question to ask yourself. First of all by definition you are only allowing finite linear combinations, that`s the main reason why you need that closure. For more on this please look at this question.
Now we can look at your question. Can you find a vector for which a finite linear combination of basisvectors would never give you the vector you`re searching for?
Hint: Look at the space $l^2$, the space of sequences $(z_n)_n$ for which $\sum_n |z_n|^2 < \infty$, notice that the "standard basis" from linear algebra, generalises to this space, and notice that the sequence $(\frac{1}{n})_n$ lives inhere. Can you write it as a finite linear combination?