What you say is correct and is true in more generality. You should be interested by the following result:
Let $K$ be a topological compact space. Then, the maximal ideals of $A
:= \mathcal{C}(K, \mathbb{R})$ are exactly the ideals of functions
vanishing at a fixed point of $K$.
Here are some hints to prove this theorem:
1) Determine the invertibles of $A$ (easy).
2) For any $x$ in $K$, show that the set of functions of $A$ vanishing in $x$ is a maximal ideal of $A$ (easy).
3) Conversely, let $I$ be a maximal ideal of $A$. Suppose by contradiction that for any $x$ in $K$, you can find a function $f_x$ in $A$ which does not vanish at $x$. By continuity, there is a neighborhood $U_x$ of $x$ such that $f_x$ does not vanish on $U_x$.
4) Using compactness, construct a function of $A$ that vanishes nowhere and conclude (clever). Hint: in the real numbers, a sum $\sum \lambda_i^2$ vanishes if, and only if, all $\lambda_i$ vanish.
Remark: adapt the proof, for functions with values in $\mathbb{C}$.
Edit: concerning the question of primality of ideals of functions vanishing at more than one point, you can prove this as follows:
Let $I_C$ be the set of functions of $A$ vanishing one some closed set $C$ of $[0,1]$. Write $C = X \cup Y$ where $X$ and $Y$ are closed proper subsets of $[0,1]$ in $C$ (this is always possible if $C$ has at least two points x < y: take for $X$ the set of points $t \in A$ with $t \leq \frac{x+y}{2}$ and for $Y$ the set of points $t \in A$ with $t \geq \frac{x+y}{2}$).
Let $f$ and $g$ be functions in $A$ vanishing exactly on $X$ and $Y$. For instance, you can take the distance functions to $X$ and $Y$. Then $f.g$ is in $I_C$ but neither $f$ nor $g$ are in $I_C$.
This can certainly be generalized for a more general topological space, but maybe the compactness condition is not sufficient.
The hint has so much implicit quantifiers, explicit them and everything becomes clear :
Assume for contradiction that the morphism $\phi : R \mapsto \mathbb{R}$ is not of ,the form $\phi_t$, ie $\boxed{\forall t \in [0;1], \, \phi \neq \phi_t}$.
For each $t \in [0;1]$, since $g_t(t)\neq 0$, there's an open neighborhood $V_t$ of $t$ such that $\forall x \in V_t, \, g_t(x)\neq 0$.
$\{V_t\}_{t \in [0;1]}$ forms an open cover of the compact set $[0;1]$, hence there is a finite subset, say $\{t_1;\dots;t_n\}$, of the index set $[0;1]$ such that $\{V_{t_1};\dots;V_{t_n}\}$ covers $[0;1]$.
Now take $g(x) := \sum_{i=1}^n g_{t_i}(x)$, etc.
Best Answer
An easy thing to prove about $C[0,1]$ (which I am using to denote the ring of continuous functions you define) is that the only idempotents are $0$ and $1$, and therefore it cannot be the product of even two nonzero rings.
As for the Chinese remainder theorem, a proof will usually go about showing that the individual coordinates (like $\{(a,0,0,0...)\mid a\in R\}$ and $\{(0,a,0,0,0...)\mid a\in R\}$ and so on) can be mapped onto, and therefore their finite sums can be mapped onto. When the index set is finite, that is the entire image, but when the index set is infinite, it is no longer the case, so the proof breaks down.