I have the following PDE
$$\frac{\partial W}{\partial s}=\frac{\partial^{2} W}{\partial x^{2}}$$ such that $W(0,s)=0$, $W(x,0)=\delta(x)$, the domain is $x>0$ and $0<u<\infty$. This question is a Dirichlet problem, so if I use the standard solution for $W(s,x)$, I should have
$$W(s,x)=\int_{0}^{\infty}\frac{1}{\sqrt{4\pi s}}[e^{-\frac{(x-y)^{2}}{4s}}-e^{-\frac{(x+y)^{2}}{4s}}]\delta(y)dy$$
This should give $W(s,x)=0$. However, at the same time, when we study heat equation with initial condition being a delta function, we always think of placing a source at 0 giving an initial impulse and let it diffuse. Our question becomes the same problem and insulating at 0. In this way, I should be able to get a solution which is not always 0, but the boundary is 0. What's wrong here? Does my conclusion $W(s,x)=0$ not correct?
It is not really clear what problem you are solving by asking that $W(x,0)=\delta(x)$, since the support of $\delta(x)$ is on the boundary where $x=0$, and you also have a Dirichlet condition there.
If you recall how your solution formula is derived, you take the odd extension of the initial data and solve the heat equation on the entire line, which ensures the boundary condition $W(0,s)=0$ holds. The problem is that the odd extension of the $\delta$ function is identically zero (this can be proved in the distributional sense). This is why you are getting $W=0$ in your formula.
Since your problem has a boundary condition (and is not posed on the whole real line), there is no fundamental solution. Instead, you have a Green's function $G(x,s;a)$, which is the solution of your half-line problem with initial data $W(x,0) = \delta(x-a)$ for $a>0$. Plugging this in you get
$$G(x,s;a) = \frac{1}{\sqrt{4\pi s}} \left( e^{-\frac{(x-a)^2}{4s}} - e^{-\frac{(x+a)^2}{4s}}\right).$$
This recovers your solution formula, which I think was your desired outcome.
In other words, for problems with boundary conditions, it is not enough to put the point source at a single point (like the origin), since the solution is not translation invariant. Instead, you have to put the point source at every point in the domain and record the impulse response. This is the Green's function method.