This question is the outcome of a few naive thoughts, without reading the proof of Gelfand-Neumark theorem.
Given a compact Hausdorff space $X$, the algebra of complex continuous functions on it is enough to capture everything on its space. In fact, by the Gelfand-Neumark theorem, it is enough to consider the commutative C*-algebras instead of considering compact Hausdorff spaces.
The important thing here is that $C*$-algebras have a complex structure. The real structure is not enough. Given the algebra $C(X, \mathbb R) \oplus C(X, \mathbb R)$ of real continuous functions on $X$, the algebra $C(X, \mathbb{C})$ is simply the direct sum $C(X, \mathbb R) \oplus C(X, \mathbb R)$, as a Banach algebra(and this can be given a complex structure, (seeing it as the complexification…)). But to obtain a C*-algebra, we need an additional C*-algebra, and the obvious way, ie, defining $(f + ig)$* $= (f – ig)$ does not work out. More precisely, the C* identity does not hold.
So one cannot weaken (as it stands) the condition in the Gelfand-Neumark theorem that we need the algebra of complex continuous functions on the space $X$, since we do need the C* structure. Of course, this is without an explicit counterexample. Which brings us to:
Qn 1. Please given an example of two non-homeomorphic compact Hausdorff spaces $X$ and $Y$ such that the function algebras $C(X, \mathbb R)$ and $C(Y, \mathbb R)$ are isomorphic(as real Banach algebras)?
(Here I am hoping that such an example exists).
Then again,
Qn 2. From the above it appears that the structure of complex numbers is involved when the algebra of complex functions captures the topology on the space. So how exactly is this happening?
(The vague notions concerning this are something like: the complex plane minus a point contains nontrivial $1$-cycles, so perhaps the continuous maps to the complex plane might perhaps capture all the information in the first homology, etc..)..
Note : Edited in response to the answers. Fixed the concerns of Andrew Stacey, and changed Gelfand-Naimark to Gelfand-Neumark, as suggested by Dmitri Pavlov.
Best Answer
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.)