 Solution

Let $m_0={\min }_{x}u(x,0)>0$, which exists by periodicity and continuity. Fix any finite interval $[0,T]$ on which the classical solution is smooth. Choose $ϵ>0$ with $ϵ(1+T)<m_0/2$ and consider the strictly positive decreasing barrier $b(t)=m_0-ϵ(1+t)$.

Initially $u>b$. If this strict inequality first failed, compactness of the spatial circle would give a first contact $(x_{\ast },t_{\ast })$ with $u=b>0$. At that contact, the spatial minimum of $u-b$ has $u_x=0$, $u_{xx}\ge 0$, and the first-contact condition gives $(u-b{)}_t\le 0$. But the equation yieldsa contradiction. Therefore $u>b$ throughout the interval. Letting $ϵ\downarrow 0$ gives $\boxed{u(x,t)\ge m_0>0}$ on $[0,T]$. Since $T$ was arbitrary, positivity persists for every time for which the assumed smooth solution exists. This argument is a parabolic comparison principle proved directly and does not assume uniform parabolicity before positivity has been established.