EDIT. The argument below is **incorrect**. Indeed, for smoothness of a divisor along $X$ one needs the zero locus of a section of $I_X/I_X^2(mH)$ to be **empty** (not just smooth), and this typically is impossible when the rank of this bundle (equal to $\mathrm{codim}(X)$) is less or equal than $\dim(X)$.

Yes, this follows from Bertini's Theorem.

Indeed, let us check that if $X \subset Y$ is an embedding of smooth projective varieties then there is a smooth divisor $D \subset Y$ containing $X$ (then we will proceed by induction). Indeed, let $I_X$ be the ideal of $X$ and let $H$ be an ample divisor class on $Y$. Then for $m \gg 0$ the sheaf $I_X(mH)$ is globally generated, hence by Bertini's Theorem on Y a general section of $I_X(mH)$ is a divisor smooth away from $X$.

On the other hand, for $m \gg 0$ we have $H^1(Y,I^2_X(mH)) = 0$, hence the morphism
$$
H^0(Y,I_X(mH)) \to H^0(X,I_X/I_X^2(mH))
$$
is surjective, and the twisted conormal bundle $I_X/I_X^2(mH)$ is globally generated. Therefore, for a general its section (hence for general section of $I_X(mH)$) the zero locus on $X$ is also smooth
(now by Bertini's Theorem on $X$), hence it is smooth everywhere.

Now, finally, we apply an inductive argument. First, we consider the embedding $X \subset X_k := \mathbb{P}^n$ and construct a smooth divisor $X_{k-1} \subset X_k$ containing $X$. Next we consider the embedding $X \subset X_{k-1}$ and construct a smooth divisor $X_{k-2} \subset X_{k-1}$ containing $X$. Iterating this procedure we construct the required chain.