Here is a slightly different, perhaps simpler take on showing that $C(X,\mathbb{R})$ determines $X$ if $X$ is compact Hausdorff. For each closed subset $K$ of $X$, define $\mathcal{I}_K$ to be the set of elements of $C(X,\mathbb{R})$ that vanish on $K$. The map $K\mapsto\mathcal{I}_K$ is a bijection from the set of closed subsets of $X$ to the set of closed ideals of $C(X,\mathbb{R})$. Urysohn's lemma and partitions of unity are enough to see this, with no complexification, Gelfand-Neumark, or (explicitly) topologized ideal spaces required. I remember doing this as an exercise in Douglas's Banach algebra techniques in operator theory in the complex setting, but the same proof works in the real setting.
Here are some details in response to a prompt in the comments. (Added later: See Theorem 3.4.1 in Kadison and Ringrose for another proof. Again, the functions are assumed complex-valued there, but you can just ignore that, read $\overline z$ as $z$ and $|z|^2$ as $z^2$, to get the real case.)
I will take it for granted that each $\mathcal{I}_K$ is a closed ideal. This doesn't require that the space is Hausdorff (nor that $K$ is closed). Suppose that $K_1$ and $K_2$ are unequal closed subsets of $X$, and without loss of generality let $x\in K_2\setminus K_1$. Because $X$ is compact Hausdorff and thus normal, Urysohn's lemma yields an $f\in C(X,\mathbb{R})$ such that $f$ vanishes on $K_1$ but $f(x)=1.$ Thus, $f$ is in $\mathcal{I}_{K_1}\setminus\mathcal{I}_{K_2}$, and this shows that $K\mapsto \mathcal{I}_K$ is injective. The work is in showing that it is surjective.
Let $\mathcal{I}$ be a closed ideal in $C(X,\mathbb{R})$, and define $K_\mathcal{I}=\cap_{f\in\mathcal{I}}f^{-1}(0)$, so that $K_\mathcal{I}$ is a closed subset of $X$. Claim: $\mathcal{I}=\mathcal{I}_{K_\mathcal{I}}$.
It is immediate from the definition of $K_\mathcal{I}$ that each element of $\mathcal{I}$ vanishes on $K_\mathcal{I}$, so that $\mathcal{I}\subseteq\mathcal{I}_{K_\mathcal{I}}.$ Let $f$ be an element of $\mathcal{I}_{K_\mathcal{I}}$. Because $\mathcal{I}$ is closed, to show that $f$ is in $\mathcal{I}$ it will suffice to find for each $\epsilon>0$ a $g\in\mathcal{I}$ with $\|f-g\|_\infty<3\epsilon$. Define $U_0=f^{-1}(-\epsilon,\epsilon)$, so $U_0$ is an open set containing $K_\mathcal{I}$. For each $y\in X\setminus U_0$, because $y\notin K_\mathcal{I}$ there is an $f_y\in \mathcal{I}$ such that $f_y(y)\neq0$. Define $$g_y=\frac{f(y)}{f_y(y)}f_y$$ and $U_y=\{x\in X:|g_y(x)-f(x)|<\epsilon\}$. Then $U_y$ is an open set containing $y$. The closed set $X\setminus U_0$ is compact, so there are finitely many points $y_1,\dots,y_n\in X\setminus U_0$ such that $U_{y_1},\ldots,U_{y_n}$ cover $X\setminus U_0$. Relabel: $U_k = U_{y_k}$ and $g_k=g_{y_k}$. Let $\varphi_0,\varphi_1,\ldots,\varphi_n$ be a partition of unity subordinate to the open cover $U_0,U_1,\ldots,U_n$. Finally, define $g=\varphi_1 g_1+\cdots+\varphi_n g_n$. That should do it.
In particular, a closed ideal is maximal if and only if the corresponding closed set is minimal, and because points are closed this means that maximal ideals correspond to points. (Maximal ideals are actually always closed in a Banach algebra, real or complex.)
Best Answer
In addition to Nik Weaver's references, let me just sketch the proof which is in fact not very difficult:
Suppose now that $\mathcal{A}$ is such a $^*$-algebra satisfying this condition for all $n$ (†).
On every finitely generated projective module $\mathcal{E}$ over $\mathcal{A}$, written without restriction as $\mathcal{E} = P \mathcal{A}^n$ with $P = P^2 = P^* \in M_n(\mathcal{A})$, the restriction $\langle . , .\rangle$ of the canonical Hermitian algebra-value inner product of the free module $\mathcal{A}^n$ is still positive (in the strongest possible sense) and non-degenerate (in the strongest possible sense that the musical homomorphism is in fact an anti-isomorphism to the dual module). This gives immediately the existence of positive algebra-valued inner products on finitely generated projective modules over such algebras.
Now suppose the algebra satisfies an additional feature: suppose $H \in M_n(\mathcal{A})$ is an invertible positive element. Then $H$ has a positive square root $H = \sqrt{H}^2$ with the property that $\sqrt{H}$ commutes with all matrices which commute with $H$ (‡). Again, a $C^*$-algebra clearly satisfies this for all $n$ by spectral calculus.
Now suppose that on the fgpm $\mathcal{E} = P \mathcal{A}^n$ you have another positive algebra-valued inner product, denoted by $h$. On the complement $P^\bot = (1 - P)\mathcal{A}^n$ we use the restriction of the canonical algebra-valued inner product to obtain a new inner product, still denoted by $h$, on the direct sum. This is still positive and has all required (very strong) non-degeneracy properties.
In conclusion: for $^*$-algebras (like e.g. $C^*$-algebras) satisfying the above two properties (†) and (‡), every finitely generated projective module carries a unique-up-to-isometry positive algebra-valued inner product.
As a side remark: many other types of $^*$-algebras satisfy these properties as well like e.g. the smooth functions on a manifold etc. So the same proof also implies (via Serre-Swan in the smooth context) that on smooth vector bundles you always have a unique-up-to-isometry positive fiber metric.
Now, to answer you question: the uniqueness gives you the desired result that every algebra-valued inner product comes from a positive fiber metric. Indeed, a positive fiber metric meets the above properties and the isometry $U$ maps fiber metrics to fiber metrics as it is algebra-linear.