I tried to prove that the set of compact operators $K(X,Y)$ is a closed subset of the set of bounded operators $B(X,Y)$ where $X,Y$ are Banach spaces. Please can you tell me if my proof is correct?
Let $u_n$ be a sequence in $K(X,Y)$ and let $u \in B(X,Y)$ be such that $\|u-u_n\| \to 0$ (in operator norm). Let $S$ denote the unit ball in $X$. The goal is to show that $u(S)$ is totally bounded:
Let $N$ be such that $\|u-u_N\|< {\varepsilon \over 2}$. Since $u_N$ is compact the image $u_N(S)$ may be covered using finitely many balls of radius ${\varepsilon \over 2}$. Let $y_1, \dots , y_K$ denote their centers. Then for every $i$, $y_i = u_N(s_i)$ for some $s_i \in S$. We claim that the balls $B(u(s_i), \varepsilon)$ cover $u(S)$. To see this let $y \in u(S)$. Then $y=u(s)$ for some $s \in S$. Let $i$ be such that $u_N(s) \in B(u_N(s_i), {\varepsilon \over 2})$. Then
$$ \|y-u(s_i)\|_Y = \|u(s)-u(s_i)\|_Y \le \|u(s) -u_N(s)\|_Y + \|u_N(s)-u_N(s_i)\|_Y < \varepsilon$$