Since this question is asked often enough, let me add a detailed solution. I'm not quite following Arturo's outline, though. The main difference is that I'm not re-proving the Cauchy-Schwarz inequality (Step 4 in Arturo's outline) but rather use the fact that multiplication by scalars and addition of vectors as well as the norm are continuous, which is a bit easier to prove.
So, assume that the norm $\|\cdot\|$ satisfies the parallelogram law
$$2 \Vert x \Vert^2 + 2\Vert y \Vert^2 = \Vert x + y \Vert^2 + \Vert x - y \Vert^2$$
for all $x,y \in V$ and put
$$\langle x, y \rangle = \frac{1}{4} \left( \Vert x + y \Vert^2 - \Vert x - y \Vert^2\right).$$ We're dealing with real vector spaces and defer the treatment of the complex case to Step 4 below.
Step 0. $\langle x, y \rangle = \langle y, x\rangle$ and $\Vert x \Vert = \sqrt{\langle x, x\rangle}$.
Obvious.
Step 1. The function $(x,y) \mapsto \langle x,y \rangle$ is continuous with respect to $\Vert \cdot \Vert$.
Continuity with respect to the norm $\Vert \cdot\Vert$ follows from the fact that addition and negation are $\Vert \cdot \Vert$-continuous, that the norm itself is continuous and that sums and compositions of continuous functions are continuous.
Remark. This continuity property of the (putative) scalar product will only be used at the very end of step 3. Until then the solution consists of purely algebraic steps.
Step 2. We have $\langle x + y, z \rangle = \langle x, z \rangle + \langle y, z\rangle$.
By the parallelogram law we have
$$2\Vert x + z \Vert^2 + 2\Vert y \Vert^2 = \Vert x + y + z \Vert^2 + \Vert x - y + z\Vert^2 .$$
This gives
$$\begin{align*}
\Vert x + y + z \Vert^2 & = 2\Vert x + z \Vert^2 + 2\Vert y \Vert^2 - \Vert x - y + z \Vert^2 \\
& = 2\Vert y + z \Vert^2 + 2\Vert x \Vert^2 - \Vert y - x + z \Vert^2
\end{align*}$$
where the second formula follows from the first by exchanging $x$ and $y$. Since $A = B$ and $A = C$ imply $A = \frac{1}{2} (B + C)$ we get
$$\Vert x + y + z \Vert^2 = \Vert x \Vert^2 + \Vert y \Vert^2 + \Vert x + z \Vert^2 + \Vert y + z \Vert^2 - \frac{1}{2}\Vert x - y + z \Vert^2 - \frac{1}{2}\Vert y - x + z \Vert^2.$$
Replacing $z$ by $-z$ in the last equation gives
$$\Vert x + y - z \Vert^2 = \Vert x \Vert^2 + \Vert y \Vert^2 + \Vert x - z \Vert^2 + \Vert y - z \Vert^2 - \frac{1}{2}\Vert x - y - z \Vert^2 - \frac{1}{2}\Vert y - x - z \Vert^2.$$
Applying $\Vert w \Vert = \Vert - w\Vert$ to the two negative terms in the last equation we get
$$\begin{align*}\langle x + y, z \rangle & = \frac{1}{4}\left(\Vert x + y + z \Vert^2 - \Vert x + y - z \Vert^2\right) \\
& = \frac{1}{4}\left(\Vert x + z \Vert^2 - \Vert x - z \Vert^2\right) +
\frac{1}{4}\left(\Vert y + z \Vert^2 - \Vert y - z \Vert^2\right) \\
& = \langle x, z \rangle + \langle y, z \rangle
\end{align*}$$
as desired.
Step 3. $\langle \lambda x, y \rangle = \lambda \langle x, y \rangle$ for all $\lambda \in \mathbb{R}$.
This clearly holds for $\lambda = -1$ and by step 2 and induction we have $\langle \lambda x, y \rangle = \lambda \langle x, y \rangle$ for all $\lambda \in \mathbb{N}$, thus for all $\lambda \in \mathbb{Z}$. If $\lambda = \frac{p}{q}$ with $p,q \in \mathbb{Z}, q \neq 0$ we get with $x' = \dfrac{x}{q}$ that
$$q \langle \lambda x, y \rangle = q\langle p x', y \rangle = p \langle q x', y \rangle = p\langle x,y \rangle,$$
so dividing this by $q$ gives
$$\langle \lambda x , y \rangle = \lambda \langle x, y \rangle \qquad\text{for all } \lambda \in \mathbb{Q}.$$
We have just seen that for fixed $x,y$ the continuous function $\displaystyle t \mapsto \frac{1}{t} \langle t x,y \rangle$ defined on $\mathbb{R} \smallsetminus \{0\}$ is equal to $\langle x,y \rangle$ for all $t \in \mathbb{Q} \smallsetminus \{0\}$, thus equality holds for all $t \in \mathbb{R} \smallsetminus \{0\}$. The case $\lambda = 0$ being trivial, we're done.
Step 4. The complex case.
Define $\displaystyle \langle x, y \rangle =\frac{1}{4} \sum_{k =0}^{3} i^{k} \Vert x +i^k y\Vert^2$, observe that $\langle ix,y \rangle = i \langle x, y \rangle$ and $\langle x, y \rangle = \overline{\langle y, x \rangle}$ and apply the case of real scalars twice (to the real and imaginary parts of $\langle \cdot, \cdot \rangle$).
Addendum. In fact we can weaken requirements of Jordan von Neumann theorem to
$$
2\Vert x\Vert^2+2\Vert y\Vert^2\leq\Vert x+y\Vert^2+\Vert x-y\Vert^2
$$
Indeed after substitution $x\to\frac{1}{2}(x+y)$, $y\to\frac{1}{2}(x-y)$ and simplifications we get
$$
\Vert x+y\Vert^2+\Vert x-y\Vert^2\leq 2\Vert x\Vert^2+2\Vert y\Vert^2
$$
which together with previous inequality gives the equality.
For part (a), you can show linearity like this:
$$
\langle\langle x + y, z \rangle\rangle = \operatorname{Re}(\langle x + y, z \rangle) = \operatorname{Re}(\langle x, z \rangle + \langle y, z \rangle) = \operatorname{Re}(\langle x, z \rangle) + \operatorname{Re}(\langle y, z \rangle) = \langle\langle x, z \rangle\rangle + \langle\langle y, z \rangle\rangle
$$
$$
\langle\langle cx, y \rangle\rangle = \operatorname{Re}(\langle cx, y \rangle) = \operatorname{Re}(c\langle x, y \rangle) = c\operatorname{Re}(\langle x, y \rangle) = c\langle\langle x, y \rangle\rangle
$$
Note that $c \in \mathbb{R}$, so the third equality in the second line is valid.
For part (b), all you need to say is that $\langle x, x \rangle \in \mathbb{R}$ even in complex inner products since it follows from the conjugate symmetric axiom. So taking the real part doesn't change the value and so $\langle\langle x, x \rangle\rangle = \langle x, x \rangle$. Now take square roots to show norms are equal.
For (c), you can clearly see that if the complex inner product is zero then taking the real part is still zero which proves the first part. But it is possible for the complex inner product to be non-zero but its real part zero (i.e. the value only has non-zero imaginary part). So the converse is false.
Best Answer
You are right.
And it has to be that way, because the norm that the Hermitian product induces on $\mathbb C^n$ is the same as the Euclidean norm on $\mathbb R^{2n}$.
Therefore, for a fixed $x\in\mathbb C^n$, the set $$ \{ y\in\mathbb C^n \mid \|x+y\|^2 = \|x\|^2+\|y\|^2 \} $$ has dimension $2n-1$ over $\mathbb R$ -- and therefore it cannot be a linear subspace over $\mathbb C$. In particular it cannot equal $\{y\in\mathbb C^n \mid \langle x,y\rangle = 0 \}$.