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.
Best Answer
Taking the squares and expanding, we get for all real number $\alpha>0$: $$\lVert x\rVert^2+\alpha\langle x,y\rangle+\alpha\overline{\langle x,y\rangle}+\alpha^2\lVert y\rVert^2\geq \lVert x\rVert^2,$$ hence $$\langle x,y\rangle+\overline{\langle x,y\rangle}+\alpha\lVert y\rVert^2\geq 0.$$ Taking $\alpha\to 0$, we get $2\Re\langle x,y\rangle\geq 0$. Working with $-x$ and $-\alpha$, we get $2\Re\langle x,y\rangle\leq 0$. For the imaginary part, work with $e^{i\theta}\beta=:\alpha$ where $\beta\in \Bbb R$ and $\theta$ such that $e^{i\theta}\langle x,y\rangle$ is a real number.