Rudin's chain rule: Why is continuity at interval necessary? Theorem 5.5, Rudin's Principles of Mathematical analysis says:

Suppose $f$ is continuous on $\color{red}{[a,b]}$,$ f'(x)$ exists at some point $x\in [a,b], g$ is defined on an interval $I$ which contains the range of $f$, and $g$ is differentiable at the  point $f(x)$. If $$h(t)=g(f(t)) \quad (a\le t \le b)$$ then $h$ is differentiable at $x$, and $$h'(x)=g'(f(x))f'(x)$$

I believe, I have understood the proof. But why is continuity at $[a,b]$ required? To me differentiablity at $x$ of $f$ and the same at $f(x)$ of $g$ is the only required conditions. 
 A: You are right, in the proof of Rudin only is used the continuity of $f$ at $x$ (consequence of differentiability without more hypothesis). In another sources (Wikipedia, Cartan...) only the differentiability of $f$ at $x$ and of $g$ at $f(x)$ is required.
A: In the first edition (1953) Rudin writes at the start of the proof:
First of all, Theorems 4.10 and 4.19 show that $R$ is an interval so that it makes sense to talk about the derivative of $g$ (we have defined the derivative only for functions defined on intervals and segments).
The premise is missing in the other editions.
Note1. The theorem assumes that "$g$ is defined on the range $R$ of $f$".
Note2. Rudin calls $[a,b]$ an interval, $]a,b[$ a segment .
A: Basically, what happens in Rudin's theorem 5.5 is this.
We start constructing an expression for a limit. 
\begin{align}
\frac{g(f(t)) - g(f(x))}{\underbrace{f(t) - f(x)}_\text{comment *}} = g'(f(x)) \quad (A)
\end{align}
Comment (*): here we use the assumption of continuity of $f(x)$, because we can find $\delta >0$ to place $t$ in the $\delta$-neighborhood of $x$. In other words, so that the denominator will converge to zero as we start to push $t$ toward $x$.
Now we construct the main expression for the limit. 
\begin{align}
\lim_{t \to x} \frac{g(f(t)) - g(f(x))}{t-x} \quad \stackrel{\text{use (A) above}}{=} \lim_{t \to x} \quad \frac{g'(f(x)) \cdot \left[ f(t) -f(x) \right]}{t-x} = g'(f(x)) \cdot f'(x)
\end{align}
as required
