Let $w_{t}$ denote the 3-dimensional heat kernel
$$w_{t}(x)=(4\pi t)^{-3/2}e^{-\left|y\right|^{2}/(4t)},\qquad y\in\mathbb{R}^{3}, \ t > 0$$ Suppose $f\in L^{3}(\mathbb{R}^{3})$, and let $u(\cdot,t)$ denote the convolution $$u(x,t)=w_{t}\ast f(x)=(4\pi t)^{-3/2}\int_{\mathbb{R}^{3}}e^{-\left|x-y\right|^{2}/(4t)}f(y)\mathrm{d}y, \qquad x\in\mathbb{R}^{3}, \ t>0$$
(I.e. $u$ is the solution at time $t$ to the linear, homogeneous equation with initial data $f$.) Show that
$$\lim_{t\downarrow 0}\sqrt{t}\left\|u(\cdot,t)\right\|_{L^{\infty}}=0 \tag{1}$$
By Young's convolution inequality, we can bound $\left\|u(\cdot,t)\right\|_{L^{\infty}}$ by $\left\|w_{t}\right\|_{L^{3/2}}$ and $\left\|f\right\|_{L^{3}}$:
$$\left\|u(\cdot,t)\right\|_{L^{\infty}}\leq \left\|w_{t}\right\|_{L^{3/2}}\left\|f\right\|_{L^{3}},\qquad\forall t>0 \tag{2}$$
By dilation invariance,
\begin{align*} \left\|w_{t}\right\|_{L^{3/2}}=(4\pi t)^{-3/2}\left(\int_{\mathbb{R}^{3}}e^{-3\left|y\right|^{2}/(8t)}\mathrm{d}y\right)^{2/3}&=\dfrac{(8t/3)\pi}{(4\pi t)^{3/2}}=\dfrac{1}{3(\pi t)^{1/2}}, \end{align*}
which yields that the quantity $\sqrt{t}\left\|u(\cdot,t)\right\|_{L^{\infty}}$ is uniformly bounded in $t>0$.
If we instead knew that $f\in L^{p}(\mathbb{R}^{3})$, for any $p>3$, then we would have the estimate
\begin{align*} \sqrt{t}\left\|u(\cdot,t)\right\|_{L^{\infty}}\leq\sqrt{t}\left\|w_{t}\right\|_{L^{p/(p-1)}}\left\|f\right\|_{L^{p}}\lesssim\sqrt{t}\left\|f\right\|_{L^{p}}\dfrac{t^{\frac{3}{2}\frac{p-1}{p}}}{t^{3/2}}=t^{\frac{1}{2}(1-\frac{3}{p})}\left\|f\right\|_{L^{3}}, \tag{3} \end{align*} which tends to zero as $t\downarrow 0$, since $1-3/p>0$.
Given this roadblock, I'm thinking that something more advanced than basic convolution estimates is needed to get decay as $t\downarrow 0$, as opposed to boundedness. Any hints or suggestions would be greatly appreciated.
I'm going to write $u_t=f*w_t$.
It seems you've shown $$\sup_{t>0}\sqrt t||u_t||_\infty\le c||f||_3$$ and $$\lim_{t\to0}\sqrt t||u_t||_\infty=0\quad(f\in L^4).$$
The result you want follows because $L^4\cap L^3$ is dense in $L^3$. By the same argument that comes up when you show boundedness of maximal functions implies almost-everywhere convegence:
Say $f\in L^3$. Let $\epsilon>0$. Choose $g\in L^4\cap L^3$ with $||f-g||_3<\epsilon$.
Write $\psi_t(h)=\sqrt t||w_t*h||_\infty$. Now
$$\limsup\psi_t(f)\le\limsup\psi_t(f-g)+\limsup\psi_t(g)=\le\sup_t\psi_t(f-g)\le c||f-g||_3\le c\epsilon,$$ so $\limsup\psi_t(f)=0$.