I came across a doubt yesterday. Is the following equality true?? $$C([0,T];C[a,b]) = C([0,T] \times [a,b])$$
I have a feeling that the answer is no. I know that if $f \in C([0,T];C[a,b])$ then $f \in C([0,T]\times [a,b])$ and in order to prove the other direction, given $f \in C([0,T]\times [a,b])$ we would have to prove that the function $g_t(x) = f(t,x)$ is such that $g_t \in C[a,b]$ for each $t$ and that $$\sup\limits_{t \in [0,T]} \|g_t\|_\infty < \infty.$$
Does anybody know an example that shows that the equality is not true?
It is true. This is a special case of the exponential law for function spaces. See for example Exponential Law in Algebraic topology. and Exponential Law (Topology) definition query. You can find it in most textbooks on general topology (provided they cover functions spaces and the compact-open topology).
However, let us deal with your question as it is. In your question all function spaces are endowed with the $\sup$-metric. I shall comment on the relation between the metric topology and the compact-open topology later.
Let me first suggest to distinguish the functions in $C([0,T];C[a,b])$ and in $C([0,T] \times [a,b])$ notationally. Given $f \in C([0,T];C[a,b])$, we write $\bar f = e(f) : [0,T] \times [a,b] \to \mathbb R$ for the function defined by $\bar f(t,x) = f(t)(x)$. You say that you already know that $\bar f$ is continuous, i.e. we have defined a function $e : C([0,T];C[a,b]) \to C([0,T] \times [a,b])$.
It remains to show that $e$ is bijective. To do so, let us consider $g \in C([0,T] \times [a,b])$. Clearly the maps $i_t : [a,b] \to [0,T] \times [a,b], i_t(x) = (t,x)$, are continuous for all $t \in [0,T]$. Hence the functions $g_t = g \circ i_t : [a,b] \to \mathbb R, g_t(x) = g(t,x)$, are continuous, i.e. we have $g_t \in C([a,b])$. This construction gives us a function $\tilde g = \bar e(g) : [0,T] \to C([a,b])$. Let us show that $\tilde g$ is uniformly continuous (this is even stronger than showing that $\tilde g$ is continuous). This means that for each $\epsilon > 0$ there exists $\delta > 0$ such that $\lvert s - t \rvert < \delta$ implies $\lVert \tilde g_s - \tilde g_t \rVert_\infty = \sup_{x \in [a,b]} \lvert g(s,x) - g(t,x) \rvert < \epsilon$. We know that $g$ is uniformly continuous, i.e. there exists $\delta > 0$ such that $\lvert s - t \rvert + \lvert x - y \rvert < \delta$ implies $\lvert g(s,x) - g(t,y) \rvert < \epsilon/2$. Hence $\lvert s - t \rvert < \delta$ implies $\lvert g(s,x) - g(t,x) \rvert < \epsilon/2$ for all $x$. We conclude $\lVert \tilde g_s - \tilde g_t \rVert_\infty \le \epsilon/2 < \epsilon$. Thus $\tilde g \in C([0,T];C[a,b]) $, i.e we have defined a function $\bar e : C([0,T] \times [a,b]) \to C([0,T];C[a,b])$.
By construction we have $\bar e \circ e = id$ and $e \circ \bar e = id$. Therefore $e$ and $\bar e$ are bijections which are inverse to each other.
What is the relation between this "metric" exponential law and the general exponential law mentioned above? This says that if $X,Y,Z$ are topological spaces and $Y$ is locally compact, then we have a bijection between the function spaces $C(X \times Y,Z)$ and $C(X,C(Y,Z))$, where $C(Y,Z)$ is endowed with the compact-open topology. But is well-known and easy to show that for a compact space $X$ and a metric space $(Y,d)$, the compact-open topology on $C(X,Y)$ agrees with the topology generated by the $\sup$-metric $d_\infty(f,f') = \sup_{x \in X} d(f(x),f'(x))$.