Recall that any metric space $M$ has a metric $d$ defined on it where:
$$d:M\times M\to \mathbb R_{\geq 0}$$
and $d$ fulfills the following axioms:
Symmetry:
$$d(x,y) = d(y,x)$$
Non-negativity
$$d(x,y)\geq 0\qquad d(x,y)=0\iff x = y$$
Triangle inequality:
$$d(x,y)\leq d(x,z)+d(z,y)$$
We can apply the triangle inequality twice to $d(x_n,y_n)$ as follows:
$$d(x_n,y_n) \leq d(x_n,y)+d(y,y_n)$$
We also have that $$d(x_n,y)\leq d(x_n,x)+d(x,y)$$
We can combine these to get that:
$$d(x_n,y_n)\leq d(x,x_n)+d(x,y)+d(y,y_n)$$
Here I implicitly used the symmetry condition to say that $d(x,x_n)=d(x_n,x)$. It wasn't required, but it's good to recognize that these are equal.
We can interchange $x_n$ and $x$ and $y_n$ and $y$ because throughout this argument we've been treating them as points, and "forgetting" that they are sequences. We could have used points called $a,b,c,d$ and still obtained a valid relation between them. So, interchange the points is just saying "this argument is true for any $4$ points, and as we don't need to repeat it, we'll just take the end result and modify it to how we want it").
Best Answer
Suppose that $x = \sum\limits_{k = 1}^\infty \alpha_k e_k$. Then for all $\epsilon > 0$, there exists $n$ s.t. $|x - \sum\limits_{k = 1}^n \alpha_k e_k| < \epsilon$. And clearly, $\sum\limits_{k = 1}^n \alpha_k e_k \in M$. Then $x$ is a limit point of $M$; that is, $x \in \overline{M}$.
Claim: for all $x$, the quantities $\alpha_1, ..., \alpha_k$ given as above minimise $|x - \sum\limits_{k = 1}^n x_k e_k|$. Proof: fairly trivial. We will use claim twice in a moment.
Conversely, suppose $x \in \overline{M}$. We wish to show that $x = \sum\limits_{k = 1}^\infty \alpha_k e_k$ where $\alpha_k$ is defined as above. Given $\epsilon > 0$, take some element $m \in M$ s.t. $|x - m| < \epsilon$ (this is possible because $x$ is a limit point of $M$). Then for some $n$, we may write $m = \sum\limits_{k = 1}^n m_k e_k$ for some complex numbers $m_k$. But it can be shown that $\alpha_1, ..., \alpha_n$ minimise $|x - \sum\limits_{k = 1}^n m_k e_k|$ Then $|x - \sum\limits_{k = 1}^n \alpha_k e_k| \leq |x - \sum\limits_{k = 1}^n m_k e_k| < \epsilon$. Now suppose that we have some $n' \geq n$. Then we see that $|x - \sum\limits_{k = 1}^{n'} \alpha_k e^k| \leq |x - \sum\limits_{k = 1}^{n'} \beta_k e_k| = |x - \sum\limits_{k = 1}^n \alpha_k e_k| < \epsilon$, where we define $\beta_k = 0$, $k > n$ and $\beta_k = \alpha_k$, $k \leq n$. We again cite our above claim to prove this.
Then by the $n-\epsilon$ definition of limit, we have $x = \lim\limits_{n' \to \infty} \sum\limits_{k = 1}^{n'} \alpha_k e_k = \sum\limits_{k = 1}^\infty \alpha_k e_k$, as required.