Let $X,Y$ be normed spaces, such that $Y$ is not complete, and $X\neq \left\{{0}\right\}$. I want to show that $L(X,Y)$ is not complete, where $L(X,Y)$ is the space of linear and bounded operators from $X$ to $Y$.
My idea of the proof is show the contrapositive. Assuming that $L(X,Y)$ is complete, let $(y_n)_{n \in{} \mathbb{N}} \subseteq Y$ be a Cauchy sequence. I want to find sequences $(x_n)_{n \in{} \mathbb{N}} \subseteq X \setminus \left\{{0}\right\}$, $(l_n)_{n \in{} \mathbb{N}} \subseteq L(X,Y)$ with $y_n=l_n(x_n)$. However for $n,m$, $ \left\|{l_n-l_m}\right\| \geq \left\|{y_n-y_m}\right\|$ , so I can't guarantee that $(l_n)_{n \in{} \mathbb{N}}$ is Cauchy.
Take $x\in X$ with $\|x\|=1$. Construct, via Hahn-Banach, a linear functional $f$ such that $f(x)=1$ and $\|f\|=1$. Finally, define linear operators $T_n\in L(X,Y)$ by $$T_n(z)=f(z)y_n.$$ Then $\|T_n(z)\|\leq\|y_n\|\,\|z\|$, and $\|T_n(x)\|=\|y_n\|$. So $\|T_n\|=\|y_n\|$. In particular, the sequence $\{T_n\}$ is Cauchy if and only if the sequence $\{y_n\}$ is Cauchy.
Now, using that $L(X,Y)$ is complete, there exists $T$ such that $T_n\to T$. Let $y=T(x)$. Then $$ \|y-y_n\|=\|(T-T_n)(x)\|\leq\|T-T_n\|\to0. $$