Weak* Convergence exercise

I'm dealing with this exercise about weak* convergence and I'm literally getting lost with indexes. I have this:

Let $$X := c_0(\mathbb{N}), \hspace{3mm}x_0 \in X^*=\ell^1(\mathbb{N}), \hspace{3mm}\{x_n\}_{n \in \mathbb{N}} \subset X^*$$ bounded. Show that

$$\begin{equation} x_n \rightharpoonup^* x_0 \Longleftrightarrow x_n(k) \rightarrow x_0(k) \end{equation}$$ for fixed $$k \in \mathbb{N}$$.

EDIT:

Hello everyone, I took back this exercise and I was trying to complete the proof, well:

1) for $$(\Rightarrow)$$ I know from HP that $$\begin{equation} \forall \{y_k\}_k \in \ell^1(\mathbb{N}) \quad, \sum_{k=1}^{\infty}x^{(n)}_ky_k -\sum_{k=1}^{\infty}x^{(0)}_k \rightarrow 0 \quad as \hspace{2mm}n \rightarrow \infty \end{equation}$$

then of course it is equivalent to say that

$$\begin{equation} \sum_{k=1}^{\infty}y_k(x^{(n)}_k-x^{(0)}_k) \rightarrow 0 \hspace{2mm}as \hspace{2mm} n \rightarrow \infty \end{equation}$$

So what I'm saying is that $$\begin{equation} \forall \epsilon \gt 0, k \in \mathbb{N}, \hspace{2mm}\exists n_{\epsilon,k} : \forall n>n_{\epsilon,k} \quad |x^{(n)}_k-x^{(0)}_k|\lt\epsilon \end{equation}$$

So I think I proved ($$\Rightarrow$$) this way (Tell me what you think)

What you are supposed to prove is the following: suppose $$\sum_j |a_j| <\infty, \sum_j |a_{nj}|$$ is bounded and $$a_{nj} \to a_j$$ as $$n \to \infty$$ for each $$j$$; then $$\sum_j a_{nj} c_j \to \sum_j a_jc_j$$ for every sequence $$(c_j)$$ which tends to $$0$$. To prove this let $$\epsilon >0$$ and choose $$N$$ such that $$|c_j| <\epsilon$$ for all $$j \geq N$$. Then $$|\sum_j a_{nj} c_j - \sum_j a_jc_j| \leq |\sum_j^{N-1} a_{nj} c_j - \sum_j^{N-1} a_jc_j|+\epsilon\sum_{j=N}^{\infty} |a_{nj}|$$. Can you now complete the proof?
[I am writing $$a_{nj}$$ for the j-th component of $$x_n$$ and $$a_j$$ for the j-th component of $$x_0$$].
Theorem. Let $$X$$ be a topological space and $$f\in C(X)$$. If $$(f_n)$$ is an equicontinuous sequence of continuous functions from $$X$$ to $$\mathbb{R}$$ that converges to $$f$$ on a dense subset, then $$(f_n)$$ converges to $$f$$ everywhere.
Now, if $$X$$ is a normed space and all the functions $$f_n$$ and $$f$$ are linear, then it suffices to have convergence on a total set (one which has a dense linear hull) instead of a dense set and equicontinuity is satisfied if $$\sup_n\lVert f_n\rVert<\infty$$. This solves the problem in question (as well as a bunch of similar exercises you might encounter in the future).