Confusion in the derivation of convolution formula $$x(t)=\int_{-\infty}^{\infty}x(\tau)\delta(\tau-t)d\tau=\int_{-\infty}^{\infty}x(\tau)\delta(t-\tau)d\tau$$
Due to linearity,
$$y(t)=T\{x(t)\}=\int_{-\infty}^{\infty}x(\tau)T\{\delta(\tau-t)\}d\tau=\int_{-\infty}^{\infty}x(\tau)T\{\delta(t-\tau)\}d\tau$$
considering the impulse response of system to be h(t),due to time invariance of unit impulse signal ,
$$y(t)=\int_{-\infty}^{\infty}x(\tau)h(\tau-t)d\tau=\int_{-\infty}^{\infty}x(\tau)h(t-\tau)d\tau$$
here, i know i have done something wrong, because 
$y(t)=\int_{-\infty}^{\infty}x(\tau)h(\tau-t)d\tau$
is the formula for cross-correlation between x(t) and h(t)
and $y(t)=\int_{-\infty}^{\infty}x(\tau)h(t-\tau)d\tau$ is the formula for convolution between x(t) and h(t)
but i cannot find, what exactly i have done wrong. could you help?
Your confusion stems from abusing the time invariance property.

Let's restate it for a simple 1D system:

$$ \text{if} ~~ T\{ x(t) \} = y(t) \implies T\{x(t-d) \} = y(t-d) $$ is correct but the following

$$ T\{ x(-t) \} = y(-t) ~~~\text{is wrong in general} $$ 

Now in your derivation you replace the input $\delta(t-\tau)$ with time reversed input $\delta(\tau-t)$ and claim the same on the output as $h(t-\tau) = h(\tau-t)$ which is not true as shown above.

The solution comes simply by recognizing the fact that $\delta(t-\tau) = \delta(-(t-\tau)) = \delta(\tau-t)$ and then proceed only with the $\delta(t-\tau)$ case as Marcus have stated.