Cofibrations are embeddings

I think I've found an answer to my question. Since the category of topological spaces is cocomplete, there exists a pushout of $A \times I \stackrel{in_0}{\longleftarrow} A \stackrel{\iota}{\longrightarrow} X$. We will denote this pushout as $M_\iota$ and it is called the mapping cylinder. We form $M_\iota$ as the a quotient space of the disjoint union of $A\times I$ and $X$ where $(a,0) \sim \iota(a)$. Since $\iota$ is a cofibration there exists a map $\phi:X \times I \rightarrow M_\iota$ s.t. $$\phi (\iota \times id) = i :A\times I \rightarrow M_f$$ Where $i$ is the inclusion of $A \times I$ into $M_f$. Thus if we identify $A \times \{1 \}$ with $A$ and $\iota(A) \times \{1 \}$ with $\iota(A)$, then $\phi$ is the inverse for $\iota$.

A variant of your proof is to consider the cone $CA = A \times I / A \times \{ 0 \}$, the cone point $\ast$ being the common equivalence class of all $(a,0)$. Let $p : A \times I \to CA$ denote the quotient map and $c : X \times \{ 0 \} \to CA, c(x,0) = *$. Then $c(\iota(a),0) = p(a,0)$, hence we get $P : X \times I \to CA$ such that $P(\iota \times id_I) = p$ and $P(x,0) = *$. But $P(\iota(a),1) = p(a,1)$ which shows as in your proof that $\iota$ is an embedding.

Note that $\iota(A)$ is not necessarily closed in $X$. Examples are inclusions $\iota : A \hookrightarrow X$, where $X$ has the trivial topology and $\emptyset \ne A \ne X$. However, if $X$ is Hausdorff, then $\iota(A)$ is closed in $X$.