Let $ S $ be a normed vector space (over $\mathbb{R}$, say, for simplicity) with norm $ \lVert\cdot\rVert$.
Can this norm be induced from inner product only through $\lVert \cdot \rVert = \sqrt{\langle \cdot, \cdot \rangle}$ ?
As to prove the if part of "A norm is induced by inner product iff the norm satisfies parallelogram equality", $\lVert \cdot \rVert = \sqrt{\langle \cdot, \cdot \rangle}$ is used.
On
In an inner product space $$ \langle v, w \rangle = \frac{1}{2}(\|v + w\|^2 -\|v\|^2 - \|w\|^2) $$ So the norm determines the inner product. The norm on a normed vector space is induced by an inner product iff the function $\langle \cdot, \cdot \rangle$ defined by the above formula satisfies the axioms for an inner product. The Jordan-von Neumann theorem says that this is so iff $\|\cdot\|$ satisfies the parallelogram identity: $$ \|v + w\|^2 + \|v - w\|^2 = 2(\|v\|^2 + \|w\|^2) $$ Normed vector spaces with this property are called Euclidean.
Well, this depends entirely on what you mean by a norm being induced by an inner product. The way this expression is usually used, the norm induced by the inner product $\langle \cdot, \cdot \rangle$ is by definition the norm $\lVert \cdot \rVert = \sqrt{\langle \cdot, \cdot \rangle}$. If you were to use the expression more loosely, you might for instance say that $\lVert \cdot \rVert = 2\sqrt{\langle \cdot, \cdot \rangle}$ is also "a norm induced by $\langle \cdot, \cdot \rangle$". You wouldn't be following standard usage, but it would be understandable. In any case, this definitional question is the only one at issue here; as far as your question seems to imply that there is anything to prove here, I believe this is mistaken. In the theorem you quote at the end of the question, the above standard definition of a norm being induced by an inner product is presupposed.