Let $X$ be a topological space and $C^0_{X,x}$ be the local $\mathbb R$-algebra of germs at $x$ of real-valued continuous maps. The maximal ideal consists of germs vanishing at $x$.
Let $A\leq C^0_{X,x}$ be a local $\mathbb R$-subalgebra. Is the inclusion map itself local?
That is, if $\mathfrak m \vartriangleleft A,\mathfrak m _x \vartriangleleft C^0_{X,x}$ are the maximal ideals, do we have $\mathfrak m=\mathfrak m_x\cap A$?
For the local $\mathbb R$-subalgebras $C^p_{X,x}\leq C^0_{X,x}$ the inclusions are indeed local. I'm not sure where to look for a counterexample.
This MO comment suggests this isn't a formal thing since the injection $\mathbb{Z}_{(p)}\rightarrow\mathbb{Q}$ is not local. (The prime maximal $(p)\mathbb Z_{(p)}\vartriangleleft \mathbb Z_{(p)}$ is non-zero but the maximal ideal of $\mathbb Q$ is zero.)
More generally, suppose $k$ is a field and $(B,\mathfrak{m})$ is a local $k$-algebra such that $B/\mathfrak{m}\cong k$ (as a $k$-algebra). Then the inclusion of any local $k$-subalgebra $A\subseteq B$ is a local homomorphism.
Indeed, observe that $A/(\mathfrak{m}\cap A)$ is naturally a $k$-subalgebra of $B/\mathfrak{m}\cong k$. The only such subalgebra is $k$ itself, so $A/(\mathfrak{m}\cap A)\cong k$. In particular, $\mathfrak{m}\cap A$ is a maximal ideal of $A$, so it must be the unique maximal ideal.