Let $(X, \mathcal{A})$ be a measurable space and $B$ a separable Banach space. Let us call a $B$-valued function $f$ on $X$ a simple function if it is of the form $\sum_{i=1}^n b_i \mathbb{1}_{A_i}$ for some $b_i \in B$ and $A_i \in \mathcal{A}$.
Without imposing any further conditions or endowing the measurable space with a measure, is there a nice characterization of the uniform closure of the class of simple functions? For example, we know that if we were dealing with real valued simple functions, then any bounded $\mathcal{A}$-measurable function could be uniformly approximated by simple functions. Obviously, one can imitate the argument and show that in the Banach-valued setting, any compact range function lies in the uniform closure of simple functions. Can we go further than that? What other functions do we have in the closure?
The range a function in the closure needs not be compact, but it needs to be totally bounded. And it is not difficult to show that the Banach space of totally bounded measurable functions (with the uniform norm) is precisely the uniform closure of the class of simple functions, if you already know how to show that any measurable function with compact range can be uniformly approximated by simple functions. And the assumption of being separable is irrelevant.
Let me add some details. First, to show that the space $\mathcal{M}_{tb}(\mathcal{A};B)$ of $B$-valued (Borel) measurable functions with totally bounded range is closed in the uniform norm topology, let $(f_{n})_{n=1}^{\infty}$ be a sequence in $\mathcal{M}_{tb}(\mathcal{A};B)$ uniformly convergent to a measurable function $f$. Recall that any sequential pointwise limit of Borel measurable functions taking values in a perfectly normal space (in particular, a metric space) should be Borel measurable. To show that the range of $f$ is totally bounded, take any $\epsilon>0$ and choose $n$ such that $\|f_{n}-f\|_{\infty}\leq\epsilon/2$. Then since the range of $f_{n}$ is totally bounded, we can find finitely many $v_{1},\ \cdots\ ,v_{K}\in B$ such that for any $x\in X$, $f_{n}(x)\in B_{\epsilon/2}(v_{k})$ for some $k=1,\ \cdots\ ,K$. Then clearly, any $f(x)$ is in some $B_{\epsilon}(v_{k})$, so the range of $f$ is indeed totally bounded.
To show that any measurable function with totally bounded range can be uniformly approximated by simple functions, take such a function $f$ and let $\epsilon>0$ be given. Then we can find $v_{1},\ \cdots\ ,v_{K}\in B$ such that $B_{\epsilon}(v_{k})$'s cover the range of $f$. Take a measurable partition $\{A_{1},\ \cdots\ ,A_{K}\}$ of $X$ such that $f(x)\in B_{\epsilon}(v_{k})$ for all $x\in A_{k}$, then the simple function $g=\sum_{k=1}^{K}v_{k}\mathbb{1}_{A_{k}}$ is uniformly approximating $f$ within $\epsilon$.