Doubt : Derive DFT from CFT
It looks like the author implicitly makes the substitution $x(nT) \rightarrow x(n)$, but even if they were making this explicit, there are more clear ways to get to Eq. 4.3.

One way to do this is to denote the continuous (analog) and discrete time-domain signals as $x_a$ and $x$, respectively. Your text, using this convention, would then look like this (starting from Eq. 4.2):

$$
x_a(t) \xrightarrow{sampling} x_a(nT)
$$

and

$$
X(e^{j\omega})\ =\ \sum_{n = -\infty}^{\infty} x_a(nT) e^{-j\Omega nT}.
$$

(Note the right-hand-side exponent: it is $-j\Omega nT$, not $-j\Omega nt$ as it appears in your text; this looks like an error to me.) Then if you take into account the following equality (assuming uniform sampling):

$$
x(n)\ =\  x_a(nT), \quad -\infty &lt; n &lt; \infty,
$$

you end up with:

$$
X(e^{j\omega})\ =\ \sum_{n = -\infty}^{\infty} x(n) e^{-j(\Omega T)n},
$$

or (noting that $\omega = \Omega T$):

$$
X(e^{j\omega})\ =\ \sum_{n = -\infty}^{\infty} x(n) e^{-j\omega n}.
$$

By the way, this is a derivation from CFT to DTFT operation, not CFT to DFT one (as you say in the title of your question).