In Rotman's Introduction to Algebraic Topology, pg 73, he shows that
If $X$ is a bounded convex subspace of a eucliden space, then $H_n(X)=0$ for all $n \ge 1$.
His proof goes as follows:
Define the cone construction homomoprhism $c_n:S_n(X) \rightarrow S_{n-1}(X)$ which satisfies $$ \partial_{n-1}c_n(\sigma) = \sigma - c_{n-1}\partial_n(\sigma).$$ for $\sigma \in S_n(X)$. This shows that $Z_n(X)=B_n(X)$.
The cone is linear extension of the following: For a fix $b\in X$, on an $n$-simplex, $\sigma:\Delta^n \rightarrow X$, we define $b \cdot \sigma : \Delta^{n+1} \rightarrow X$ where $$ b \cdot \sigma:= \begin{cases} b& \text{ if } t_0=1 \\ t_0b+(1-t_0)b(\frac{t_1}{1-t_0}, \ldots, \frac{t_{n+1}}{1-t_0} ) & \text{ if } t_0 \not= 1 \end{cases} $$
all that is required for the operation to be well defiend is that of convexity. Where is boundedness required?
A convex subset of a Euclidean space is a star domain (with respect to any of its points), therefore it is contractible, from which it follows that its homology groups are all trivial.
The boundedness condition is either a typo or an artifact related to the definitions encountered thus far in the book.