Fundamental solution of Heat equation For every bounded function $u_0\in C(\mathbb R)$ the function $$u(t,x)=\frac{1}{\sqrt{4\pi t}}\int_{-\infty}^{\infty}e^{-\frac{-(x-y)^2}{4t}}u_0(y)dy$$ is a solution of $\dfrac{\partial v}{\partial t}(t,x)=\dfrac{\partial^2v}{\partial x^2}(t,x)$
and $\lim\limits_{t \to 0}u(t,x)=u_0(x)$.
To prove that is it just necessary to differentiate $$\frac{1}{\sqrt{4\pi t}}e^{-\frac{-(x-y)^2}{4t}}u_0(y)\;?$$ Because when I differentiate it one time to $t$ and two times to $x$ the results are not the same (they differ in the sign). Is there a better method to prove it is a solution?
That $\lim\limits_{t \to 0}u(t,x)=u_0(x)$ is from the physical point of view clear, from the mathematical not when I let $t \to 0$ in the integral term. 

Differentiating the expression for $u(t,x)$ should work, just write it out again more carefully and you should find your mysterious negative sign.  You can also do this "symbolically" by letting $H(t,x)=\frac{1}{\sqrt{4\pi t}}e^{-x^2/4t}$, verifying that $H$ solves $u_t=u_{xx}$ (by taking derivatives carefully) and writing the integral for $u(t,x)$ as: 
$$
u(t,x)=\int_{-\infty}^\infty H(t,x-y)u_0(y)dy
$$
so that 
$$
u_t=\int H_t(t,x-y)u_0(y)dy\\
u_{xx}=\int H_{xx}(t,x-y)u_0(y)dy\\
\Rightarrow u_t-u_{xx}=\int (H_t(t,x-y)-H_{xx}(t,x-y))u_0(y)dy=0
$$ since $H_t=H_{xx}$.
As for showing rigorously that $\lim_{t\rightarrow 0}u(t,x)=u_0(x)$, this requires quite a bit more effort. I would consult a book on basic Fourier analysis (like Stein and Shakarchi) or PDE - the solution is too long to write out here.  You have to make a change of variables, then break the integral up into different parts and carefully control the size of each part in order to show that it goes to $u_0$.  Essentially what you're doing is showing that this function $H$ (the "heat kernel") is an approximation to the identity, or in other words, $\lim_{t\rightarrow 0}H(t,x)=\delta_0(x)$, the Dirac delta distribution. 
