How to prove that square-summable sequences form a Hilbert space?

Let $\ell^2$ be the set of sequences $x = (x_n)_{n\in\mathbb{N}}$ ($x_n \in \mathbb{C}$) such that $\sum_{k\in\mathbb{N}} \left|x_k\right|^2 < \infty$, how can I prove that $\ell^2$ is a Hilbert space (with dot-product $\left(x,y\right) = \sum_{k\in\mathbb{N}} x_k\overline{y_k}$). This is a standard textbook exercise: apparently this is easy and, even to me, it seems self-evident. However, I don't know what to do with the infinite sum.

You need to show that $\ell^2$ is a vector space, that the dot product satisfies the properties of inner product, and that the space is complete relative to the induced metric

The tricky bit in showing it is a vector space is showing that it is closed under sums; this is a consequence of the Hölder inequality.

Showing that the dot product is an inner product is straightforward: you may need to use the Hölder inequality or Cauchy-Schwarz inequality to show that $\sum x_k\overline{y_k}$ converges when both $(x_n)$ and $(y_n)$ are square summ

And then you have the issue of showing that $\ell^2$ is actually complete relative to the induced norm. This is pretty standard (in fact, one shows that the spaces $\ell_p$ are Banach spaces for all $p\geq 1$, hence complete, so $\ell^2$ is complete.

Added. From the comments, it seems your main difficulty is operating with the infinite sums and making sure they have the properties you want, similar to finite sums. (Let me commend you for your caution: it is indeed important not to simply assume that properties that hold for finite sums will transfer to infinite series). We use the fact that, by definition, $$\sum_{k=1}^{\infty} a_k = \lim_{n\to\infty}\sum_{k=1}^n a_n.$$ In the sense that the series on the left converges to the limit, if the limit exists.

So, for example, you have: \begin{align*} \overline{\langle (x_{n}),(y_{n})\rangle} &= \overline{\sum_{k=1}^{\infty}x_{k}\overline{y_{k}}}\\\ &=\overline{\lim_{n\to\infty}\sum_{k=1}^{n} x_{k}\overline{y_{k}}}\\\ &= \lim_{n\to\infty}\overline{\sum_{k=1}^{n} x_{k}\overline{y_{k}}}&&\mbox{(since conjugation is continuous)}\\\ &= \lim_{n\to\infty}\sum_{k=1}^{n}\overline{x_{k}\overline{y_{k}}} &&\mbox{(since the sum is finite)}\\\ &= \lim_{n\to\infty}\sum_{k=1}^{n}\overline{x_{k}}\overline{\overline{y_{k}}}\\\ &= \lim_{n\to\infty}\sum_{k=1}^{n}\overline{x_{k}}y_{k}\\\ &= \lim_{n\to\infty}\sum_{k=1}^{n}y_{k}\overline{x_{k}}\\\ &= \sum_{k=1}^{\infty} y_{k}\overline{x_{k}}\\\ &= \langle (y_{k}),(x_{k})\rangle. \end{align*} Etc.

Show that your definition $\left(x,y\right) \equiv \sum_{k\in\mathbb{N}} x_k\overline{y_k}$ Satisfies all the inner product axioms.

Hint: axioms are

1. conjugate symmetric
2. linearity in one of the arguments
3. Positive definite: $(u,u)\geq 0$ equality iff u=0

Now show that $F(x) = (x,z)$ defines a bounded linear functional $F$ on your inner product space $X$ of norm $||z||$. If the mapping $z\rightarrow F$ ( $X\rightarrow X'$ where X' is space of continuous linear functionals) is surjective, it must be a hilbert space.

EDIT OK, since you're worried about infinites let me share something I discovered recently in a book called the Cauchy Schwarz masterclass, the author calls it the principle of qualitative inference.

Given: $\sum_{k\in\mathbb{N}} a_k^2 \lt \infty$ and $\sum_{k\in\mathbb{N}} b_k^2 \lt \infty$ then it can be shown that $\sum_{k\in\mathbb{N}} a_k b_k \lt \infty$

For large k

$a_k b_k \leq \frac{1}{2} (a_k^2+b_k^2)$

$\Rightarrow \sum_{k\in\mathbb{N}} a_k b_k \leq \sum_{k\in\mathbb{N}} \frac{1}{2} (a_k^2+b_k^2)$

This already gives you a glimpse at the truth of the assertion.

(On a sidenote, you can go on to introduce $\hat{a}_k=\frac{a_k}{\sqrt{\sum_{k\in\mathbb{N}} a_k^2}}$ and derive the Cauchy inequality!)

Now do for conjugate symmetric what was done for real symmetric.