If you have an inner product space $\left(E, \varphi\right)$, it has a natural structure as a normed vector space: $\left(E,x\mapsto \sqrt{\varphi(x,x)}\right)$ but the other way around isn't true. There are norms that do not come from inner products.
And example with $E=\Bbb R^2$
If you take $\varphi:\left(\left(x_1,y_1\right),\left(x_2,y_2\right)\right)\mapsto x_1x_2+y_1y_2$ you have an inner product.
And if you let $N_2:\left(x,y\right) \mapsto \sqrt{\varphi\left(\left(x,y\right),\left(x,y\right)\right)}=\sqrt{x^2+y^2}$, you get the norm you know.
But there are other norms such as $N_\infty:(x,y)\mapsto \max(x,y)$ that can't be built from an inner product.
By the way, if your norm $N$ does come from an inner product, you can get the inner product back by letting $\psi:(x,y)\mapsto \cfrac{N(x+y)
^2-N(x-y)^2}{4}$
If you have a vector space $X$ with an inner product $\langle \cdot, \cdot \rangle$, this defines a norm $\|\cdot\|$ by $\|x\|=\sqrt{\langle x, x\rangle}$ (it is a good exercise to prove that this is in fact a norm). Similarly, this defines a metric, $d(x,y)=\|x-y\|$ (it is again a good exercise to prove that this is in fact a metric). This is the case for any inner product space, so yes, an inner product always defines a metric. However, not every metric is defined by an inner product!
A sequence of elements $\{x_n\}$ in $X$ is called a Cauchy sequence if $\|x_n-x_m\|\to0$ as $n,m\to\infty$. An inner product space $X$ is called a Hilbert space if it is a complete metric space, i.e. if $\{x_n\}$ is a Cauchy sequence in $X$, then there exists $x\in X$ with $\|x-x_n\|\to0$ as $n\to\infty$.
Best Answer
Yes, every inner product space is a metric space, with the "Euclidean metric" defined by $$ d(x,y) = \sqrt{\langle x-y, x-y \rangle} $$
Not every metric on a vector space comes from an inner product though (For instance, $l^1$, the space of summable sequences, is one such example)