Let $F$ be a Hausdorff topological field. Let us require all algebras to be associative commutative unital, and algebra morphisms to be unital.
Let us say a sheaf $\mathcal{O}$ of $F$-algebras on a topological space $X$ has property $\mathcal{P}$ if: for every open $U\subset X$ and $x \in U$ and $f \in \mathcal{O}(U)$, there exists a unique $\lambda \in F$ satisfying the property that $f-\lambda$ is non-invertible when restricted to any open neighborhood of $x$ (contained in $U$).
Could anyone help me verify if the following proof is correct?
Claim: the sheaf of continuous $F$-valued functions on $X$ has property $\mathcal{P}$.
Proof: Given $U,f,x$ as above;
Existence: let $\lambda = f(x)$; then $f-\lambda$, restricted to any open neighborhood of $x$, has a zero at $x$ and therefore cannot be invertible.
Uniqueness: suppose $\alpha \in F$ is such that $f-\alpha$ is non-invertible in any neighborhood of $x$; it follows that $f-\alpha$ has a zero in any neighborhood of $x$; thus $x$ is a limit point of $f^{-1}(\{\alpha\})$, but the latter is closed since $f$ is continuous and $F$ is Hausdorff, hence $x \in f^{-1}(\{\alpha\})$, hence $\alpha = f(x)$.
Edit: generalized from $\mathbb{R}$ to any Hausdorff topological field $F$.
Edit: as pointed out below, more generally $F$ need only be $T_1$ (which is weaker than Hausdorff), i.e. all its points are closed.
Your proof looks correct - in fact, we only need the field’s topology to be $T_1$.
Existence proceeds exactly as you described. For uniqueness, consider some $\beta \neq f(x)$. Then let $V = f^{-1}(F \setminus \{\beta\})$, which is open since $F$ is $T_1$ and $f$ is continuous. Then $f - \beta$ is nonzero everywhere on $V$, hence invertible there. And $x \in V$.