Let $$ v(x,t)=\frac{1}{\sqrt{4\pi t}}\int_\mathbb{R} e^{-(x-y)^2/(4t)}u_0(y)\, dy $$ be the fundamental solution of the heat equation.
Would like to prove that $$ \lvert v(x,t)-v(x,t')\rvert\leq \left\lVert\frac{d^2}{dx^2}u_0\right\rVert_{L^\infty}\lvert t-t'\rvert. $$
The only way I see is to use integration by parts.
Substituting $z=\frac{y-x}{2\sqrt{t}}$, I get
$$ v(x,t)=\frac{1}{\sqrt{\pi}}\int_\mathbb{R}e^{-z^2} u_0(2\sqrt{t}z+x)\, dz $$
Integrating by parts twice gives me $$ v(x,t)=\left[\frac{\textrm{erf}(z)}{2}\cdot u_0(2\sqrt{t}z+x)\right]_{-\infty}^\infty - \sqrt{t}\left[\left(z\cdot\textrm{erf}(z)+\frac{e^{-z^2}}{\sqrt{\pi}}\right)\cdot u_0'(2\sqrt{t}z+x)\right]_{-\infty}^\infty+2t\int_\mathbb{R} \left(z\cdot\textrm{erf}(z)+\frac{e^{-z^2}}{\sqrt{\pi}}\right)u_0''(2\sqrt{t}z+x)\, dz $$ and, analogously, $$ v(x,t')=\left[\frac{\textrm{erf}(z)}{2}\cdot u_0(2\sqrt{t'}z+x)\right]_{-\infty}^\infty - \sqrt{t'}\left[\left(z\cdot\textrm{erf}(z)+\frac{e^{-z^2}}{\sqrt{\pi}}\right)\cdot u_0'(2\sqrt{t'}z+x)\right]_{-\infty}^\infty+2t'\int_\mathbb{R} \left(z\cdot\textrm{erf}(z)+\frac{e^{-z^2}}{\sqrt{\pi}}\right)u_0''(2\sqrt{t'}z+x)\, dz $$
Looks terrible. However, can one use this to get the desired result?
I have not tried to go through the entire argument myself, but have you considered using the fact that
$$|v(x, t)- v(x, t')| = |\int_{t'}^t v_{\tau}(x, \tau) \, d \tau| = |\int_{t'}^t v_{xx} (x, \tau) \, d \tau| \leq \lVert v_{xx}(x, \tau) \rVert_{\infty} |t-t'|$$
and try to relate $\lVert v_{xx}(x, \tau) \rVert_{\infty}$ to $\lVert u_0 ''(x) \rVert_{\infty}$?