You are misunderstanding Sternberg's argument. The argument goes like this (with my notations):

Consider two representations $V$ and $W$ of $G$ with the same character.

For any representation $S$ of $G$, let $\chi_S$ denote the character of $S$.

Take *any* decomposition $V=V_1\oplus V_2\oplus ...\oplus V_k$ of $V$ into irreducible subrepresentations.

Take *any* decomposition $W=W_1\oplus W_2\oplus ...\oplus W_l$ of $W$ into irreducible subrepresentations.

For any irreducible representation $P$ of $G$, we have $\left(\chi_P,\chi_V\right) = \left(\text{the number of }i\in\left\lbrace 1,2,...,k\right\rbrace \text{ such that }V_i\cong P\right)$ and $\left(\chi_P,\chi_W\right) = \left(\text{the number of }j\in\left\lbrace 1,2,...,l\right\rbrace \text{ such that }W_j\cong P\right)$. Since $\chi_V = \chi_W$, we thus have

$\left(\text{the number of }i\in\left\lbrace 1,2,...,k\right\rbrace \text{ such that }V_i\cong P\right)$

$= \left(\chi_P,\chi_V\right) = \left(\chi_P,\chi_W\right)$

$= \left(\text{the number of }j\in\left\lbrace 1,2,...,l\right\rbrace \text{ such that }W_j\cong P\right)$

for every irreducible representation $P$ of $G$. In other words, for every irreducible representation $P$ of $G$, the two lists $\left(V_1,V_2,...,V_k\right)$ and $\left(W_1,W_2,...,W_l\right)$ contain the same amount of entries isomorphic to $P$. In other words, the lists $\left(V_1,V_2,...,V_k\right)$ and $\left(W_1,W_2,...,W_l\right)$ contain the same entries *up to isomorphism* the same number of times (but of course, not necessarily in the same order). As a consequence of this, $V_1\oplus V_2\oplus ...\oplus V_k \cong W_1\oplus W_2\oplus ...\oplus W_l$. In other words, $V\cong W$, qed.

Nowhere did this use that the $V_i$ are somehow unique. (They are not unique as subspaces of $V$. They are unique up to isomorphism, and that follows either from the Krull-Remak-Schmidt theorem or from the fact that

$\left(\text{the number of }i\in\left\lbrace 1,2,...,k\right\rbrace \text{ such that }V_i\cong P\right) = \left(\chi_P,\chi_V\right)$.)

the isomorphism classes of the $w_j$ are unique. But it's not necessary to know this in order to follow Sternberg's argument. It is actually a corollary of Sternberg's argument. $\endgroup$ – darij grinberg Jul 8 '12 at 11:12