About the closedness of $\frac d{dx}$ operator
Let us begin by recalling that an operator $A:X \to X$ with domain $D(A) \subseteq X$ is closed if its graph $$G(A) = \{(x, Ax) : x \in D(A)\} \subseteq X \times X$$ is closed.
For the first example, this means we want to check that if $f_n \in D(A) = C^1[a,b]$ and $f_n \to f$, $Af_n = f_n' \to g$ in $X = C[a,b]$ then we have that $f \in C^1[a,b]$ and $f' = g$. This is really a simple exercise is real analysis. Start by writing $$f_n(t) = f_n(a) + \int_a^t f_n'(s) ds$$ and then send $n \to \infty$ (using the fact that the convergence $f_n' \to g$ is uniform to pass the limit through the integral) to see that $$f(t) = f(a) + \int_a^t g(s) ds.$$ So $f \in C^1[a,b]$ and $f' = g$ as desired.
For the second example, it suffices to take some sequence of functions $f_n \in C^\infty[a,b]$ such that $f_n \to f$, $f_n' \to f'$ in $C[a,b]$ with $f \not \in C^\infty[a,b]$ (since then $(f_n, f_n')$ is a sequence in $G(A)$ that converges in $X \times X$ to an element of the complement of $G(A)$). To do this, you can use the fact that $C^\infty$ is dense in $C^1$ with its usual norm (simply pick any $f$ in $C^1 \setminus C^\infty$ and approximate it in $C^1$-norm by elements of $C^\infty$). This idea also tells you that any closed extension of this second operator must contain the first operator.
Hence, since the operator $A$ in the first example is a closed extension of the one in the second, it follows that this second operator is closable (since that just means that it has a closed extension) and that its closure is the operator of the first example.