Let $X$ be a topological space. For each open $U\subset X$ let $\mathcal{F}(U)$ be the ring of real functions on $U$ (not necessarily continuous). Is this a sheaf?
If $\mathcal{F}(U)$ is the ring of continuous functions from U to a topological space Y, this is a sheaf. But in the proof that this is a sheaf, where do we use continuity?
Yes, the presheaf of all functions to $\Bbb R$ a sheaf. The glueing property of sheaves is clearly satisfied.
For the sheafiness of continuous functions, the key is to note that a function $f:U\to Y$ is continuous iff it is locally continuous, that is if for all $x\in U$ there is an open neighbourhood $V$ of $X$ where $V\subseteq U$ with $f|_V$ continuous. One can clearly always glue a compatible family of continuous functions to make a locally continuous function. As locally continuity implies continuity, that ensures that the continuous functions form a sheaf.