Consider the map $\mathbb{R} \to C(0,1)$ which sends $a \in \mathbb{R}$ to the constant function with value $a$. Suppose $m \subset C(0, 1)$ is a maximal ideal with the property that the composition $\mathbb{R} \to C(0,1) \to C(0,1)/m$ is an isomorphism. Show that $m$ is of the form $\{f\in C(0,1)|f(p)=0\ \ \text{for some}\ \ p\in (0,1)\}$.
Since the composition map is bijective, therefore the map $C(0,1)\to C(0,1)/m$ must be surjective. This implies that the map is the canonical map. Now, how to proceed further. Thanks beforehand.
Obviously, any function in $m$ must have a root. As the residual field is $\mathbb{R}$, there is $a \in \mathbb{R}$ such that $x-a \in m$ (where $x$ is the identity $(0,1) \rightarrow \mathbb{R}$). As $x-a$ must have a root in $(0,1)$, $a \in (0,1)$.
Now, let $f \in m$. Then $g=f^2+(x-a)^2 \in m$, so we cannot have $g > 0$. But if $t \neq a$, $g(t) \geq (t-a)^2 > 0$. So $g(a) \leq 0$ which implies $f(a)=0$.
Thus $m \subset \{f \in C((0,1)),\, f(a)=0\}$. But the RHS is an ideal as well, so the inclusion is an equality.