"We know that a distribution with zero Skewness are symmetric." This is not the case. A symmetric distribution has zero skewness, but a distribution can have zero skewness and be asymmetric.
The skewness of a random variable $X$ is defined as
$$\gamma_1 = \mathrm{E}\left[\left(\frac{X-\mu}{\sigma}\right)^3\right],$$
where $\mu=\mathrm{E}[X]$ and $\sigma = \sqrt{\mathrm{E}[(X - \mu)^2]}$.
Notice, it's the first odd central moment of the distribution, normalized to the variance (the variance is the first even central moment).
More terminology: a distribution's moments are defined by
$$E[X^n] = \int x^n f(x) \mathrm{d}\,x$$
for $f$ the probability density function of the random variable $X$. A central moment is one where the mean has been shifted away, that is
$$E[(X-\mu)^n] = \int (x-\mu)^n f(x) \mathrm{d}\,x.$$
A moment is odd or even depending on if $n$ is odd or even.
Now, the mean is the first odd moment of the distribution, right? Crucially, if a distribution is even as a function about a point, then that point has to be the function's mean and median. Why is that? Well, if we integrate an odd function on an interval that is symmetric about the point the function is odd across, then we get zero. If $f$ is even about some point of symmetry $x_s$, then the quantity $(x-x_s)f(x)$ will be odd about that point. That is enough to prove that $x_s$ is the mean of the distribution (algebra left for the reader).
Showing that the median of a symmetric distribution is at the point of symmetry is fairly straightforward - the definition of the median is that half of the probability is on one side of the point, half of the other. If a function is symmetric then the integral of the function on one side of the point of symmetry has to be the same as the integral on the other (assuming the integration regions are symmetric, to).
Now, showing that the point of symmetry is not necessarily the mode is best done with an example. Consider the random variable with the pdf
$$f(x) = \frac{1}{2\sqrt{2\pi}} \left(e^{-(x+2)^2/2} + e^{-(x-2)^2/2}\right).$$
The distribution is symmetric about $x=0$, but the distribution has a minimum at $x=0$, not a maximum. So, you know that the point of symmetry is a minimum or maximum, because its derivative has to vanish there (why?), but it could be a local min or local max, instead of a global max.
Constructing a distribution with vanishing skewness that is asymmetric would require a little more work. Start with the standard normal distribution
$$f_N(x) = e^{-(x-\mu)^2/2\sigma^2}.$$
Now, perturb it by multiplying by (1+ax^2). You'll find that to normalize the new pdf you need to divide it by
$$N_{\mathrm{new}} \sqrt{2\pi}\sigma(a \sigma^2 + a\mu^2 + 1),$$
and the new mean is
$$\mu_{\mathrm{new}} = \mu \frac{3 a \sigma^2 + a\mu^2 + 1}{a\sigma^2 + a\mu^2 + 1}.$$
If you compute the third central moment you'll find that you can make it vanish when
\begin{align}
a & = 0 \text{ or} \\
a & = \frac{3}{\mu^2 - 3\sigma^2}.
\end{align}
Now, we need $a\ge0$ for $f$ to be positive semi-definite, so the existence of a real solution will depend on whether $\mu > \sqrt{3}\sigma$ or not. The $a=0$ solution is the trivial one where the distribution is symmetric about the mean, so it doesn't pass the test of showing an asymmetric distribution with vanishing skewness.
Appendix:
A function is even about a point $x_s$ if it satisfies
$$f([x-x_s] + x_s) = f(-[x-x_s]+x_s)$$
and it is odd about $x_s$ if
$$f([x-x_s] + x_s) = -f(-[x-x_s]+x_s).$$
