Let $V$ be a real affine variety with its ring $\mathbb{R}[V]$ of regular functions. Denote by $|\mathbb{R}[V]|$ the set of all $\mathbb{R}$-algebra homomorphisms $h:\mathbb{R}[V]\to \mathbb{R}.$
Any point $x\in V$ induces element $ev_x\in|\mathbb{R}[V]|$ beeing the evaluation map at $x$. I.e. $ev_x(f)=f(x).$
Question. Do all elements from $|\mathbb{R}[V]|$ are of form $ev_x$?
Suppose $V \subset \mathbb{A}_{\mathbb{R}}^n$, and consider the ideal $I(V) \subset \mathbb{R}[x_1, \cdots, x_n]$. Then $\mathbb{R}[V] \cong \mathbb{R}[x_1, \cdots, x_n]/I(V)$. Let $\overline{f}$ denote the image of $f\in \mathbb{R}[x_1, \cdots, x_n]$ in $\mathbb{R}[V]$.
Since $\mathbb{R}[V]$ is generated by $\overline{x_k}$ for $1\le k \le n$, any $\mathbb{R}$-algebra homomorphism $\phi: \mathbb{R}[V]\to \mathbb{R}$ is entirely determined by the images of the $\overline{x_k}$.
If $\phi(\overline{x_k}) = c_k$, then for any regular function $\overline{f}(x_1, \cdots, x_n)\in \mathbb{R}[V]$, we have $$\phi(\overline{f}(x_1, \cdots, x_n)) = f(\phi(\overline{x_1}), \cdots, \phi(\overline{x_n})) = f(c_1, \cdots, c_n)$$
Set $x := (c_1, \cdots, c_n) \in \mathbb{A}_{\mathbb{R}}^n$. Notice that for any $g\in I(V)$, we have $g(x) = \phi(\overline{g}) = \phi(0) = 0$. Thus, in fact $x \in V$.
We conclude $\phi = \text{ev}_x$ for $x\in V$.