Let $\mathfrak{X}$ be a normed space and let $\mathfrak{Y}$ be a complete normed space. Prove that $\mathfrak{L}(\mathfrak{X},\mathfrak{Y})$ is complete.
As far as I'm aware, proving a space is complete requires proving that every Cauchy sequence converges but I'm unsure on how to do this.
If $T_n$ is a Cauchy sequence in $\mathcal L (X,B)$, then for any chosen $x \in X$, $T_n(x)$ is a Cauchy sequence in $B$. Since $B$ is complete, this sequence converges.
Define a new linear operator $T : X \to B$ mapping $x \mapsto \lim_{n \to \infty} T_n(x)$.
Now prove that $T_n \to T$ in the operator norm. Let $\epsilon > 0$. The Cauchy property tells you that there exists an $N$ such that $$ m,n > N \implies \sup_{||x || \leq 1} ||T_n(x) - T_m (x) || < \epsilon.$$ Take the limit $m \to \infty$.
[If your notation $\mathcal L (X, B)$ refers to bounded operators, then you also need to prove that $T$ is bounded. I'll leave you to do that if required.]