If $E_1$ of a finite dimensional real inner product space, then every coset of $E/E_1$ contains exactly one vector orthogonal to $E_1$ In the book of linear algebra by Werner Greub, at page $191$ it is asked that

Given a subspace $E_1$ of a finite dimensional real inner product
  space $E$, consider the factor space $E/E_1$.Prove that every
  equivalence class, i.e coset, contains exactly one vector which is
  orthogonal to $E_1$.

As a proof method, I thought I should first show that every coset contains a orthogonal vector to $E_1$, and then show that this orthogonal vector is unique to the coset.I tried the prove the first step by assuming that there is some coset in $E/E_1$ s.t it doesn't contain a orthogonal vector to $E_1$, but couldn't derive anything useful.
I have also considered the linear function space of $E$, i.e the isomorphism between them, but I couldn't find anything there either.
 A: For existence:
If the coset is $E_1$ itself, then ofcourse $0$ is the unique vector that is orthogonal to $E_1$. If it is not $E_1$ then it is of the from $v+E_1$ for some $v\notin E_1$. Take the projection $Pr_{E_1}(v)$ of $v$ onto $E_1$. Then $v-Pr_{E_1}(v)$ is orthogonal to $E_1$ and still an element of the coset (as $-Pr_{E_1}(v)\in E_1$).
For uniqueness:
Take two such vectors in $v + E_1$, i.e. $v+x \in E_1^\perp$ and $v+y \in E_1^\perp$, where $x,y\in E_1$. Since $E_1^\perp$ is itself a subspace we have that the diffrence of the two vectors is still in there:
$$(v+x)-(v+y) = x-y \in E_1^\perp$$
but $x-y$ is also an element of $E_1$ so it must be $0$.
A: Hint: Select an orthonormal basis $\{e_1,e_2,\dots,e_d\}$ of $E_1$.  A vector is orthogonal to $E_1$ if and only if it satisfies
$$
\langle x, e_i \rangle = 0  \qquad i = 1,\dots,d
$$
Now, a coset can be written in the form
$$
\{x_0 + e : e \in E_1\}
$$
and we see that the desired conclusion is equivalent to saying that the system of equations
$$
\langle x_0 + e, e_i \rangle = 0  \qquad i = 1,\dots,d \implies\\
\langle e, e_i \rangle = - \langle x_0, e_i \rangle  \qquad i = 1,\dots,d 
$$
(where $x_0$ is a fixed element of $E$)
has a unique solution $e \in E_1$.  To put it another way: it is equivalent to show that the operator $f:E_1 \to \Bbb R^d$ given by
$$
f(e) = (\langle e, e_1 \rangle,\langle e, e_2 \rangle,\dots,\langle e, e_d \rangle)
$$
is an isomorphism.

Even if $\{e_1,\dots,e_d\}$ is an arbitrary basis (not necessarily orthonormal), all of the steps in the above "hint" apply.  The slightly tricky aspect now, however, is showing that $f$ is an isomorphism.
Because the spaces are finite dimensional, it suffices to note that $\dim(E_1) = \dim(\Bbb R^d)$, and to show that $\ker(f) = \{0\}$.
Now, suppose that $x \in \ker f$. That is, $\langle x, e_i \rangle = 0$ for all $i$.  In order to show that $x = 0$, it suffices to show that $\langle x, y \rangle = 0$ for all $y \in E$.  To that end, note that for any $y \in E$, we can select coefficients $\alpha_i$ such that
$$
y = \sum_{i=1}^d \alpha_i e_i
$$
Conclude that 
$$
\langle x,y \rangle = \left \langle x, \sum_{i=1}^d \alpha_i e_i \right \rangle = 
\sum_{i=1}^d \alpha_i \langle x, e_i\rangle = 0
$$
