Theorem: Let $(Y,d)$ be a complete metric space. Then the metric space $(\mathcal{B}([a,b], Y), d^{\sup})$ is complete, where $B(C,D)$ is the set of all bounded functions form $C$ to $D$ and $d^{\sup}(f,g) := \sup_{x \in [a,b]} d(f(x), g(x))$ for $f,g: [a,b] \to Y$ is the supremum-metric induced by $d$.
I now want to find a simple counterexample showing this implication doesn't hold the other way around but haven't been able to get any help from this related question.
The converse also holds:
For any nonempty set $X$, we have an isometric embedding $c:Y\hookrightarrow \mathcal B(X,Y)$, taking the constant maps.
Now, if $(y_n)$ is a Cauchy sequence, then so is $c(y_n)$, so it converges to some $f\in\mathcal B(X,Y)$.
Then, by definition of $d^\sup$, evaluating at any $x\in X$, we get $y_n\to f(x)$.