Suppose you have a metric space $(X,d)$ and a point $x\in X$ such that every ball of finite radius $B(x;r)$ is contractible. Then is $X$ itself contractible?
Edit: I've been asked to provide context for this question, to save it from being closed. Well, the question just occurred to me out of the blue with no particular application in mind (although generally speaking I'm not interested in anything more exotic than a CW-complex). It seems to me a naturally interesting question whether contractibility of finite balls implies anything about the contractibility of a potentially infinite space $X$. Certainly when $X$ is a finite metric space the answer is yes. And a little thought shows that the converse statement is false in general. So what remained was to understand the case of infinite metric spaces, and Qiaochu's answer below provides the answer for (spaces with homotopy type of) CW-complexes, which is good enough for me. (As for the "level" of the question being asked, the fact that I accepted the answer should indicate something about that.)
$X$ must be weakly contractible in the sense that all of its homotopy groups (at every basepoint) vanish, as follows: let $f : S^n \to X$ be a continuous map from a sphere into $X$. Since $S^n$ is compact, the image of $S^n$ is bounded, and so contained in some ball of finite radius, which we can take WLOG to be centered at any particular basepoint. Since this ball is contractible, the image of $S^n$ can be contracted to a point. So $f$ is null-homotopic wrt any basepoint in the image of $f$.
By Whitehead's theorem it follows that if $X$ has the homotopy type of a CW complex then it is contractible. I don't know how to remove this hypothesis.