Solution for Cauchy Problem $u_t-u_{xx} = 0$ belongs to the Gevrey class of order $1/2$ 
Let $u(x,t)$be the solution for the Cauchy Problem
$$u_t-u_{xx} = 0 \mbox{ in $\mathbb{R}\times ]0, \infty[$}$$ $$u(x,0)
 = u_0(x) \mbox{ in $\mathbb{R}$} $$
where $u_0\in S(\mathbb{R})$(schwartz space in $\mathbb{R})$. Conclude that, for each fixed $t_0>0$,
  the function $x\to u(x,t_0)$ belongs to the Gevrey class of order
  $1/2$ in $\mathbb{R}$

I already defined the Gevrey class here: $\sup_K |\partial^{\alpha}u|\le C^{|\alpha|+1}\alpha!^s$ then $u$ is analytic for $s\le 1$ but here is the definition again:

A function $u\in C^{\infty}$ belongs to the Gevrey Class of order $s$
  if for every compact $K$ of $\Omega$ there is a constant $C$ such that
$$\sup_K |\partial^{\alpha}u|\le C^{|\alpha|+1}\alpha!^s, \ \alpha\in\mathbb{Z}_+^N$$

this exercise comes after this one:

define, for $x\in\mathbb{R}$, 
$$v(x) = \int_{\mathbb{R}}e^{ix\lambda-a\lambda^2}d\lambda$$
Show that $v$ belongs to the Gevrey cass of order $1/2$ in
  $\mathbb{R}$

So maybe they have something in common. The last exercise above looks like a fourier transform. 
UPDATE
$$||\partial_x^n \phi||_{L^2} = (\frac{1}{\sqrt{4\pi t}}\int|\partial_{x}^n| e^{-(x\sqrt{4t})^2}dx)^{1/2} = (\frac{1}{\sqrt{\pi}}\int|\partial_x^n e^{-x^2}|dx)^{1/2} = ||\partial_x^n \phi_{1/4}||L^2$$
 A: In this answer, all norms are spatial norms (i.e. there is no integration/supremum wrt the parameter $t_0$ that was already fixed in the question, and I have dropped the subscript $0$).
The unique solution to the heat equation with nice initial data is well known as the convolution with the Gaussian kernel,
$$ u(x,t) = \phi_t * u_0(x) = \frac1 {\sqrt{4\pi t}}\int_{\mathbb R} e^{-|x-y|^2/4t} u_0(y) \ dy.$$
Then by Cauchy-Schwarz ($|\int fg|\le \|f\|_{L^2}\|g\|_{L^2}$ ),
\begin{align}
&\|\partial_x^n u(x,t)\|_{L^\infty} \\ 
& = \|(\partial_x^n \phi_t) * u_0\|_{L^\infty} \\
& = \left \|\int_{\mathbb R}\partial^n_x\phi_t(x-y) u_0(y) \ dy\right \|_{L_x^\infty} \\
& \le  \left \| \|\partial^n_x\phi_t(x-y)\|_{L^2_y} \|u_0\|_{L^2}\right \|_{L_x^\infty} \\
& = \left \| \|\partial^n_x\phi_t\|_{L^2} \|u_0\|_{L^2}\right \|_{L_x^\infty} \\
& =  \|\partial^n_x\phi_t\|_{L^2} \|u_0\|_{L^2} \\
&\le C^{n+1} \|\partial_x^n \phi_{1/4}\|_{L^2} \|u_0\|_{L^2} \\
&\le C_1^{n+1} \|\partial_x^n \phi_{1/4}\|_{L^2} \end{align}
Above, we used Chain rule,
$$\partial^n_x\phi_t(x) = C(t) \partial^n_x \exp(-x^2/(4t)) = \frac{1}{(\sqrt{4t})^{n}} (\partial_x^n\phi_{1/4})(x/\sqrt{4t}) $$
and also $ \int_{\mathbb R} f(\lambda x)^2 dx = \frac1\lambda \int_{\mathbb R}  f^2(y) dy $, so that
$$ \|\partial^n_x\phi_t\|_{L^2} \le C^{n+1}\|\partial_x^n \phi_{1/4}\|_{L^2} $$
Note that $$\partial_x^n \phi_{1/4}(x) =  (-1)^nH_n(x) \phi_{1/4}(x),$$ where $H_n$ is the $n$th (physicist) Hermite polynomial.
The Hermite polynomials have the well known $L^2$ norm when weighted with $\phi_{1/4}$ (which I quote from that same wikipedia page),
$$\int H_n^2 \phi_{1/4}^2 \le \|\phi_{1/4}\|_{L^\infty} \int H_n^2 \phi_{1/4} \le C \sqrt{2\pi} n! $$
Thus
$$ \|\partial_x^n u(x,t)\|_{L^\infty} \le C_2^{n+1} \sqrt{n!}.$$

Response to update in OP: note that you seem to have avoided chain rule by using
$$ \int |\partial_x^n [\phi_{1/4} (\lambda x) ]|^2 \ dx \overset{?}{=} \int |[\partial_x^n \phi_{1/4}] (\lambda x)|^2  \ dx  $$
Note the very intentional bracketing. This isn't true, try writing it out explicitly for $n=1$.
A: By your second box, we have the result that

If $v_a$ is defined by
  $$v_a(x) = \int_{\mathbb{R}} e^{ix\lambda -a\lambda^2}\;d\lambda$$
  the $v_a$ is in the Gervy class of order $\frac{1}{2}$. 

In fact, if you look at the result from Finding $\sup_{\lambda \ge 0}{\lambda^k e^{−a\lambda^2/2}}$, you see that we actually have something better:
$$\sup_{x \in \mathbb{R}} \partial^\alpha v(x) \le C^{|\alpha| + 1} \alpha!^\frac{1}{2}$$
Moreover, $v_t$ is actually (up to a scaling factor) the fundamental solution of the heat equation. We can obtain it by taking the Fourier transform in $x$ with frequency variable $\lambda$, solving the resulting ODE, and inverting the transform.
Thus, it would suffice to prove the following result:

$$\sup_{x} \partial^\alpha v_t * g (x) \le C^{\alpha + 1}\alpha!^\frac{1}{2}$$ for any Schwartz class $g$. 

Note that
$$\begin{align*}
\sup_{x} |\partial^\alpha (v_t * g)(x)| &=\left\lVert(\partial^\alpha v_t) * g (x)\right\rVert_{L^\infty}\\
 &\le \lVert \partial^\alpha v_t\rVert_{L^\infty} \lVert g \rVert_{L^1}\\
 &\le C^{|\alpha|+1}\lVert g \rVert_{L^1} \alpha!^\frac{1}{2}
\end{align*}$$
by Holder's inequality.  It follows that $u(x,t) = v_t * g$ is in the Gervy class of order $\frac{1}{2}$ for any fixed $t > 0$.
