An arbitrary subspace inclusion need not be a cofibration

Take $X=\{0\}\cup \{1\mathrm{/}n:n\ge 1\}\subset \mathbb{R}$ and A={0}. The complement of A is open in X, since each point 1/n is isolated, so A is closed. Every path in X is constant: if two of its values differ, an irrational strictly between them would be a value of the composite real-valued path on the subinterval between those times by F3, but X contains only rational numbers.

If A→X had HEP, use target $S=X\times \{0\}\cup \{0\}\times I$, initial map x↦(x,0), and homotopy (0,t)↦(0,t). F1 would give $R:X\times I\to S$ fixing those prescribed points. Its first coordinate along {1/n}×I is a path in X starting at 1/n and therefore constant by step 1.1. The only point of S with first coordinate 1/n is (1/n,0); hence $R(1\mathrm{/}n,1)=(1\mathrm{/}n,0)$ for every n.

But $(1\mathrm{/}n,1)\to (0,1)$ in $X\times I$. Continuity would make the second coordinates of their R-images converge to the second coordinate of $R(0,1)=(0,1)$, namely 1. They are all zero by step 2.1, a contradiction. Thus this explicit closed inclusion is not a cofibration.