The function $$f(x,t)=\begin{cases} \frac{1}{\sqrt t}e^{-x^2/4t} ~~;& t>0, ~x\in\mathbb R \\ ~~~~~~0 ~~~~~~~~;& t\le 0,~ x\in\mathbb R \end{cases}$$is a solution of the heat equation in
$1.~~~\{(x,t):x\in\mathbb R,~t\in\mathbb R\}$
$2.~~~\{(x,t):x\in\mathbb R,~t>0\}$ but not in the set $\{(x,t):x\in\mathbb R,~t<0\}$
$3.~~~\{(x,t):x\in\mathbb R,~t\in\mathbb R\}$ \ $\{(0,0)\}$
$4.~~~\{(x,t):x\in\mathbb R,~t>-1\}$
Clearly, the two first order partial derivative $~u_x~$ and $~u_t~$ do not exist at $(0,0)$, so I think only option $\bf{(3)}$ is correct.
However I think that there is another well-explained answer of this question. If anybody have it, please provide to us.
The kernel of the heat equation i.e. the soution of the PDE $$\begin{cases} \frac{\partial f}{\partial t}=c \frac{\partial^2 f}{\partial x^2},\quad (x,t)\in \mathbb{R}\times (0,\infty)\\ f(x,0)=\delta(x) \end{cases}$$ is the function defined on the region $\mathbb{R}\times (0,\infty)$(without the constants).
If we consider the function with domain $\mathbb{R}\times \mathbb{R}$ we can see that the function tell us how a constant function $0$ evolves acording to the Heat equation given that, at some time we apply a "perturbation" (in this case on $t=0$). For example the solution $f=0$ means that no external "perturbation" was made.
Now, in your question, if we want a solution as you impose for $t>0$ the PDE that $f$ must satisfy is $$\begin{cases} \frac{\partial f}{\partial t}=c \frac{\partial^2 f}{\partial x^2},\quad (x,t)\in \mathbb{R}\times (-n,\infty)\\ f(x,0)=\delta(x) \end{cases}$$
In particular at $t=0$ f itself it's not a function, it is a distribution. On the other hand you defined $f(x,0)=0$ so both conditions $(f(x,0)=0$ and $f(x,0)=\delta(x))$ are mutually exclusive (for $x=0$) and therefore, the only way to define a function is to remove the point ${(0,0)}$.