I am struggling with this simple example of a compact operator.
Let $c_0=\{x\in l^{\infty}(\mathbb{N})|x=(x_i) \text{ and } \lim_{i\to \infty} x_i=0\}$ endowed with the norm $||x||_{\infty}=\sup_i |x|_i$. For given $x\in c_0$ let $f_x:c_0\to c_0$ be given by $$ (y_i)_{i\in \mathbb{N}} \mapsto ((x_i^2)y_i)_{i\in \mathbb{N}}.$$ Show that $f_x$ is compact.
I showed $f_x$ is well-defined, linear and bounded. I assume $(y)$ is a bounded sequence in $c_0$ (so any sequence in $c_0$). I want to find a subsequence $(y^{n_k})_i$ such that $||f_xy^{n_k}_i-f_xy_i|| \to 0$ and I note that $$||f_xy^{n_k}_i-f_xy_i||=||(x^i)^2(y_i^{n_k}-y_i)||\leq M^2||y_i^{n_k}-y_i||$$ as $\sup_i|x_i|=M<\infty$. So I want $||y^{n_k}_i-y_i||\to 0$. But now as $y\in c_0$, $y_i\to 0$ and $y^{n_k}_i\to 0$, $||y_i||+||y^{n_k}_i||\to0$ the claim would be true for any subsequence which doesn't seem right.
This is just a sketch of proof, but I hope it can help you.
Let $N_k$ such that $\forall n, x_{N_k+n}^2\leq \frac1k$. You can extract a subsequence of $y$ such that it's $N_k$ first terms are distant of at most $\frac1k$. You have: $$ \|f_x y^{n_i}-f_x y^{n_j}\|\leq \max\left(\underbrace{\frac{\|x^2\|}{k}}_{\text{The first terms}},\underbrace{\frac{2\|y\|}{k}}_{\text{The last terms}}\right). $$ Indeed, for $l\leq N_k$ we have: $$ |f_x y^{n_i}-f_x y^{n_j}|_l\leq \|x^2\|\max_{0\leq m\leq N_k}|y^{n_i}-y^{n_j}|_m\leq \frac{\|x^2\|}{k}, $$ and for $l\geq N_k$ we have: $$ |f_x y^{n_i}-f_x y^{n_j}|_l\leq \|y^{n_i}-y^{n_j}\|_\infty\sup_{N_k\leq m}|x_m^2|\leq \frac{2\|y\|}{k}. $$
Using Cantor's diagonal extraction, you have the desired result.