Number of orthonormal sets of vectors $\leq$ dim of the vector space 
Theorem. If $V,W$ are arbitrary representations of $G$, say
  $$V=V_1^{a_1}\oplus \dots \oplus V_k^{a_k} $$
  $$W=V_1^{b_1}\oplus \dots \oplus V_k^{b_k} $$
  Then,
  $$\langle \chi_V,\chi_W \rangle = \sum_{i=1}^k a_ib_i$$
Corollary. Number of irreducible representations of $G \leq$ number of conjugacy classes of $G$.
Proof.
  Number of orthonormal sets of vectors $\leq$ dim of the vector space
  $$\chi_1,\dots,\chi_k \leq \dim \mathbb{C}^{\mathcal{C}}=|\mathcal{C}| $$

I cannot see why this proof works. Im guessing that the LHS of the inequality is the rows of the character table. I cannot see exactly why these would be orthonormal. Moreover I cannot see how $\dim \mathbb{C}^{\mathcal{C}}$ is related to conjugacy classes.
Please do not use the theorem which for me comes later that states equality.
 A: Well for the dimension thing you must see that if $C=\{c_1,...,c_r\}$ :
$$\mathbb{C}^C=\{\text{functions from } C\text{ to } \mathbb{C}\}=\{(f(c_1),...,f(c_r))\in\mathbb{C}^r\}"="\mathbb{C}^r $$
Where $r$ is the number of elements in $C$. So this is kind of trivial.
Now to proove the corollary. From the theorem you get that two different irreducible representations leads to orthogonal characters. Take $i\neq j$ then :
$$V_i=0V_1\oplus...\oplus 0V_{i-1}\oplus 1V_i\oplus 0V_{i-1}\oplus...\oplus 0V_k$$
$$V_j=0V_1\oplus...\oplus 0V_{j-1}\oplus 1V_j\oplus 0V_{j-1}\oplus...\oplus 0V_k$$
Now (from the theorem) you have that :
$$<\chi_i,\chi_j>=\sum_{l=1}^k\delta_{i,k}\delta_{j,k}=0\text{ because } i\neq j $$
So if you take $V_1,...,V_k$ all the irreducible representations then :
$$(\chi_1,...,\chi_k)\text{ is an orthonormal family for the above scalar product} $$
In particular $(\chi_1,...,\chi_k)$ is free (i.e. there cannot be any non-trivial linear combination of this vectors). And because it is a family included in $\mathbb{C}^C$ you have that :
$$k\leq dim(\mathbb{C}^C)=|C| $$
