The completion of a normed linear space is a Banach space, and similarly inner product spaces (pre-Hilbert spaces) complete to Hilbert spaces. The key to this is the following important fact about completions:
Suppose $(M_1, d_1)$ and $(M_2, d_2)$ are metric spaces, with completion $(\overline{M}_1, \overline{d}_1)$ and $(\overline{M}_2, \overline{d}_2)$ (see footnote). If $f : M_1 \to M_2$ is a uniformly continuous function, then there exists a unique continuous extension $\overline{f} : \overline{M}_1 \to \overline{M}_2$ of $f$.
Let's assume this fact, because this answer will be sufficiently long, and see how we use it. Let's suppose that $X$ is a normed linear space. First, we need to establish vector addition and scalar multiplication on $\overline{X}$. The map $+ : X \times X \to X$ is uniformly continuous where $X \times X$ has the metric
$$d((a, b), (c, d)) = \|a - c\| + \|b - d\|,$$
which is one of the infinitely many metrics that produce the product topology. To see this, note that
$$\|+(a, b) - +(c, d)\| = \|a + b - c - d\| \le \|a - c\| + \|b - d\| = d((a, b), (c, d)).$$
Thus the $+$ map is non-expansive, and hence uniformly continuous. We can therefore extend it to a map $\oplus : \overline{X \times X} \to \overline{X}$ that is continuous. We also need to note the fact that $\overline{X \times X}$ is homeomorphic to $\overline{X} \times \overline{X}$ in the expected way.
We can also do something similar to the scalar multiplication map $\cdot : \Bbb{F} \times X \to X$ to produce a map $\odot : \Bbb{F} \times X$. Unfortunately, the original map is not quite uniformly continuous, but it is uniformly continuous on bounded sets, so we can just restrict to each bounded set, and complete the map on each such set. Because $\Bbb{F} \times X$ is connected, the completion is unique, and Cauchy sequences representing points in $\overline{X}$ can be assumed to be as bounded as you like, the argument still carries through, and we do still get a unique continuous map $\odot$.
These functions are addition and scalar multiplication operations on $\overline{X}$, and satisfy the various axioms of vector spaces, which you can prove by continuity. For example, consider the distributive law:
$$a \odot (x \oplus y) = a \odot x \oplus a \odot y.$$
We may choose $x_n \to x$ and $y_n \to y$ where $x_n, y_n \in X$, and using the corresponding property in $X$,
$$a \cdot (x_n + y_n) = a \cdot x_n + a \cdot y_n$$
for all $n$. We can write this in $\overline{X}$, remembering that $x_n, y_n \in X \subseteq \overline{X}$, so
$$a \odot (x_n \oplus y_n) = a \odot x_n \oplus a \odot y_n.$$
Since $\odot$ and $\oplus$ are continuous, we can take the limit of both sides with respect to $n \to \infty$, and the distributive law is proven. You can probably appreciate how most other axioms follow in this vein.
The additive identity will just be $0 \in X \subseteq \overline{X}$. Additive inverses are similarly easy: $-[(x_n)] := [(-x_n)]$. Closure under addition and scalar multiplication, if your personal definition of vector spaces explicitly includes such a thing, are guaranteed by the codomains of $\oplus$ and $\odot$.
So, with a lot of work, we can show $\overline{X}$ is at least a vector space. To show its a Banach space, you can extend $\|\cdot\|$ in the same way. You can then show the axioms of a norm using continuity as before, and then show that it agrees with the metric on $\overline{X}$. This shows that $\overline{X}$ is a normed linear space, and because of the final point, $\overline{X}$ is complete under this norm.
To get the final claim, you guessed it, we complete the inner product map $\langle \cdot, \cdot \rangle : X \times X \to \Bbb{F}$. Like the scalar multiplication, this is uniformly continuous on bounded sets, but not uniformly continuous otherwise. One can show that the axioms of an inner product still hold, and that its relationship to the previously constructed norm still hold, by continuity.
There's a lot to prove here, but I think you have enough to point you in the right direction.
