Is the proof of Lemma A.2.3.1 in HTT correct? I am reading the proof of proposition A.2.3.1 in Lurie's HTT, the proposition can be stated as the following:

Let $\mathscr{C}$ be a a model category, $A$ and $B$ be cofibrant objects, and $X$ be a fibrant object. Let $f: A \to X$ be a map, and $i: A \to B$ be a cofibration. If there is a map $g: B \to X$ such that $g \circ i$ is homotopic to $f$, then there is a map $g': B \to X$ such that $g' \circ i = f$ and $g'$ is homotopic to $g$.  

The proof goes as follows:

Let $C(A)$ be a cylinder object for $A$, and consider a factorization $C(A) \coprod\limits_{A \coprod A}\left(B \coprod B\right) \to C(B) \to B$ be a factorization of $C(A) \coprod\limits_{A \coprod A}\left(B \coprod B\right) \to B$ as a cofibration followed by a trivial fibration. By hypothesis, $g'$ and $f \circ i$ are homotopic so there is a map $h_0: C(A) \coprod_A B \to X$ such that ${h_0}_{|B} = g'$ and ${h_0}_{|A} = f$. The inclusion $C(A)\coprod_A B \to C(B)$ is a trivial cofibration and so there is an extension of $h_0$ to a map $h: C(B) \to X$ which gives the desired homotopy as $C(B)$ is a cylinder object for $B$.

I am fine with most of the proof, except that I do not see why $C(A)\coprod_A B \to C(B)$ is a trivial cofibration. Or rather, I am unsure wether it is clear yet:
I know that since $A$ is cofibrant, the inclusions $A \to C(A)$ are trivial cofibrations, and since $A \to B$ is a cofibration, the pushout $B \to C(A)\coprod_A B$ is a weak equivalence, as well as a cofibration, The composite $B \to C(B) = B \to C(A)\coprod_A B \to C(B)$ is as well an acyclic cofibration. And so by $2$ out of $3$ $C(A)\coprod_A B \to C(B)$ must be a weak equivalence as well, and it is a cofibration.
My problem with this is: this uses the fact that pushout of a weak equivalence along a cofibration between cofibrant objects is a weak equivalence, and this fact is proved only later in Lurie's HTT, and the proof relies on proposition A.2.3.1, so this would be circular.
I know there are other ways to prove A.2.3.1, by considering path objects insteaf of cylinders objects, as it is done in Hirschorn's book, but I wonder wether I am actually missing something in Lurie's proof and wether there is an other way of proving that $C(A)\coprod_A B \to C(B)$ is a weak equivalence that doesn't rely on the fact that a pushout of weak equivalences along a cofibration between cofibrant objects is a weak equivalence.
 A: The situation which needs considering in this case is not the pushout of a weak equivalence along a cofibration, but rather the pushout of a trivial cofibration along an arbitrary map. To this end let $i:A\xrightarrow\simeq B$ be both a cofibration and a weak equivalence, let $f:A\rightarrow X$ is an arbitrary map, and consider the pushout square
$\require{AMScd}$
\begin{CD}
    A@>f>>X\\
    @ViV\simeq V   @VV j V\\
    B @>g>> B\cup_AX.
\end{CD}
Recall that the trivial cofibrations in a model category are characterised as those morphisms which have the left lefting property (llp) with respect to all fibrations. Therefore, to test whether the cobase change $j$ is an acyclic cofibration it suffices to find a diagonal lift in any diagram of the form
$\require{AMScd}$
\begin{CD}
   X@>\alpha>>E\\
       @VV j V@VVpV\\
     B\cup_AX@>\beta>>F.
\end{CD}
where $p:E\rightarrow F$ is a fibration. To solve this problem consider the compound diagram 
$\require{AMScd}$
\begin{CD}
    A@>f>>X@>\alpha>>E\\
    @ViV\simeq V   @VV j V@VVpV\\
    B @>g>> B\cup_AX@>\beta>>B.
\end{CD}
Since $i$ is a trivial fibration, a map $k:B\rightarrow E$ can be found satisfying $ki=\alpha f$ and $pk=\beta g$. Since the left-hand square is a pushout, there then exists a map $l:B\cup_AX\rightarrow E$ with $lg=k$ and $lj=\alpha$. Since $plj=p\alpha =\beta j$ and $plg=pk=\beta g$, we conclude from the uniqueness granted by the univerisal property of the pushout that $pl=\beta$. Hence $l$ is a solution to the lifting problem, and we may conclude that $j$ is a trivial fibration.
