I am currently reading a textbook in which it is claimed that both the following maps are (Hurewitz) cofibrations $A\vee A\rightarrow A\wedge (I_+)$ (the first $A$ is mapped to $A\times\{0\}$ and the second to $A\times\{1\}$) and $A\rightarrow CA$, where $CA$ is the reduced cone on $A$, and $A$ is a non-degenerately based space.
I have thought about showing that these subspaces are NDR using the DR of $A\wedge (I_+)$ to $A$, but my attempts have not been successful so far.
You need some knowledge about cofibrations. I don't know which book you use, but most authors focus on closed cofibrations. These are cofibrations $i : A \to X$ such that $i(A)$ is closed in $X$. Also NDR pairs $(X,A)$ are usually only defined for closed $A \subset X$. The definition of NDR pairs can be generalized to arbitrary $A \subset X$, but we do not consider this here.
We need the following facts about cofibrations.
(1) Let $i : A \to X$, $f : A \to B$, $i' : B \to Y$ and $f' : X \to Y$ be four maps which form a pushout diagram. If $i$ is cofibration, then also $i'$ is a cofibration.
(2) An inclusion $i : A \to X$ of a closed subspace is a cofibration if and only if $(X,A)$ is a NDR pair.
(3) If $(X,A), (Y,B)$ are NDR pairs, then so is $(X \times Y, X \times B \cup A \times Y)$.
$A$ has a non-degenerate basepoint $a_0$ which means that $(A, \{ a_0 \})$ is an NDR pair.
$A \vee A$ is obtained from the disjoint union of two copies of $A$ by identifying the two copies of $a_0$. An equivalent representation is $$A \vee A = (A \times \{ 0, 1\} \cup \{ a_0 \} \times I) / \{ a_0 \} \times I .$$ Let $p : A \times \{ 0, 1\} \cup \{ a_0 \} \times I \to A \vee A$ denote the quotient map.
$A \wedge I_+$ is nothing else than $A \times I / \{ a_0 \} \times I$. Let $p' : A \times I \to A \times I / \{ a_0 \} \times I$ denote the quotient map. Its restriction $p''$ to $A \times \{ 0, 1\} \cup \{ a_0 \} \times I$ induces a unique map $i' : A \vee A \to A \times I / \{ a_0 \} \times I$ such that $i' \circ p = p'' = p' \circ i$ where $i : A \times \{ 0, 1\} \cup \{ a_0 \} \times I \to A \times I$ denotes the inclusion map. Obviously $A_k = p(A \times \{ k \}) \subset A \vee A$ is mapped by $i'$ homeomorphically onto $A'_k = p'(A \times \{ k \}) \subset A \times I / \{ a_0 \} \times I$.
It is an easy exercise to verify that the four maps $i, p, i', p'$ form a pushout diagram.
By (3) $(A \times I, A \times \{ 0, 1\} \cup \{ a_0 \} \times I)$ is an NDR pair. Therefore $i$ is a cofibration and (1) shows that also $i'$ is one.
The reduced cone $CA$ can be represented as
$$CA = (A \times I / \{ a_0 \} \times I) / A'_1 .$$
Let $r' : A \times I / \{ a_0 \} \times I \to CA$ denote the quotient map. The map $i'' : A \to CA, i''(x) = r'(p'(x,0))$, is the canonical embedding of $A$ as the base of $CA$.
The map $R : A \times \{ 0, 1\} \cup \{ a_0 \} \times I \to A, R(x,t) = x$ for $t = 0$, $R(x,t) = a_0$ for $t > 0$, induces a unique map $r : A \vee A \to A$ such that $R = r \circ p$. It is the identity on $A_0$ and contracts $A_1$ to $a_0$.
Again it is an easy exercise to verify that the four maps $i', r, i'', r'$ form a pushout diagram. This implies that $i''$ is a cofibration since $i'$ is one.