This is a proof-verification request.
Let $(X,d)$ be a compact metric space and let $\mathbf{C}(X)$ denote the set of (bounded and) continuous real-valued functions on $X$. Suppose that $\mathcal A\subseteq \mathbf{C}(X)$ is a subalgebra that contains the constant functions and separates points; that is:
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$f,g\in\mathcal A$ and $\alpha,\beta\in\mathbb R$ imply $\alpha f+\beta g\in\mathcal A$;
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$f,g\in\mathcal A$ imply that $fg\in\mathcal A$;
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all constant functions on $X$ are included in $\mathcal A$; and
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$x,y\in X$ and $x\neq y$ imply the existence of some $f\in\mathcal A$ such that $f(x)\neq f(y)$.
The Stone–Weierstrass theorem implies that $\mathcal A$ is dense in $\mathbf{C}(X)$ (with respect to the supremum metric $d_{\infty}$).
I propose a result and a proof showing that one can do even better: approximating any continuous function on $X$ uniformly by members of $\mathcal A$ while reaching exact agreement at any finite number of points!
This result was put forth by Boel et al. (2001) for a more general case (locally compact Hausdorff spaces and complex-valued continuous functions vanishing at infinity), but I felt more comfortable with constructing a slightly different proof.
Proposition: Suppose that $\varphi\in\mathbf{C}(X)$ and that $x_1,\ldots,x_k$ are distinct points in $X$ ($k\in\mathbb N$). Then, for any $\varepsilon>0$, there exists some $f\in\mathcal A$ such that $d_{\infty}(\varphi,f)<\varepsilon$ and $\varphi(x_i)=f(x_i)$ for all $i\in\{1,\ldots,k\}$.
Before proving this proposition, an auxiliary result is needed.
Lemma: For any $(c_1,\ldots,c_k)\in\mathbb R^k$, there exists some $g\in\mathcal A$ such that $g(x_i)=c_i$ for $i\in\{1,\ldots,k\}$.
Proof of lemma: Define the evaluation functional $\Phi:\mathbf{C}(X)\to\mathbb R^k$ as $$\Phi(\varphi)\equiv(\varphi(x_1),\ldots,\varphi(x_k))$$ for every $\varphi\in\mathbf C(X)$. Letting $$\delta\equiv\min_{\substack{i\in\{1,\ldots,k\}\phantom{\setminus\{i\}}\\j\in\{1,\ldots,k\}\setminus\{i\}}}d(x_i,x_j)>0,$$ one can define $$\varphi(x)\equiv\sum_{i=1}^kc_i\max\left\{1-\frac{d(x,x_i)}{\delta},0\right\}$$ for every $x\in X$. It is not difficult to check that $\varphi\in\mathbf{C}(X)$ and $\varphi(x_i)=c_i$ for every $i\in\{1,\ldots,k\}$. This entails that $\Phi$ is surjective: $\Phi(\mathbf{C}(X))=\mathbb R^k$. The Stone–Weierstrass theorem, in turn, implies that $\Phi(\mathcal A)$ is a dense linear subspace of $\mathbb R^k$. Since $\Phi(\mathcal A)$ is finite-dimensional, it must be closed, so that $\Phi(\mathcal A)=\mathbb R^k$. The result follows. $\blacksquare$
Proof of proposition: I am going to proceed by induction on the number of points. The case $k=1$ is easy: take any $h\in\mathcal A$ such that $d_{\infty}(\varphi,h)<\varepsilon/4$ and denote $\Delta\equiv\varphi(x_1)-h(x_1)$. Defining $$f(x)\equiv h(x)+\Delta$$ for every $x\in X$, one has that $f\in\mathcal A$. Now, for any $x\in X$, $$|f(x)-\varphi(x)|\leq|h(x)-\varphi(x)|+|h(x_1)-\varphi(x_1)|\leq 2d_{\infty}(\varphi,h)<\frac{\varepsilon}{2},$$ so that $d_{\infty}(f,\varphi)\leq\varepsilon/2<\varepsilon$. Clearly, $f(x_1)=\varphi(x_1)$ by construction.
Now suppose that the result holds up until $k\in\mathbb N$. For $k+1$, choose some $g\in\mathcal A$ such that
\begin{align*}
g(x_1)=\cdots=g(x_k)=&\;0,\\
g(x_{k+1})=&\;1,
\end{align*}
which is possible by the lemma. Let $$\xi\equiv\sup_{x\in X}|g(x)|\in(0,\infty).$$ By the induction hypothesis, it is possible to choose $h\in\mathcal A$ in such a way that $$d_{\infty}(\varphi,h)<\min\left\{\frac{\varepsilon}{4},\frac{\varepsilon}{4\xi}\right\}$$ and $h(x_i)=\varphi(x_i)$ for $i\in\{1,\ldots,k\}$. Let $\Delta\equiv\varphi(x_{k+1})-h(x_{k+1})$ and define $$f(x)\equiv h(x)+\Delta g(x)$$ for every $x\in X$. Clearly, $f\in\mathcal A$, $f(x_i)=\varphi(x_i)$ for every $i\in\{1,\ldots,k,k+1\}$, and for any $x\in X$, $$|f(x)-\varphi(x)|\leq|h(x)-\varphi(x)|+|g(x)||\Delta|\leq d_{\infty}(h,\varphi)+\xi d_{\infty}(h,\varphi)<\frac{\varepsilon}{2},$$ which implies that $d_{\infty}(f,\varphi)\leq\varepsilon/2<\varepsilon$. The induction thus goes through from $k$ to $k+1$ and the proof is complete. $\blacksquare$
Best Answer
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Yes, your proof is correct.