Why atmospheric pressure does not influence vapor pressure of water but only temperature does?

But why is [the vapour pressure] not dependent on atmospheric pressure or pressure applied on the gas surrounding vapour-water equilibrium?

Does this beg the question? In other words, is the vapor pressure really independent of the surrounding pressure? Pressurizing a system tends to increase its energy, which we'd expect to affect the vapor pressure as an indication of the state of the condensed matter beneath it. Let's see.

The vapor pressure can be modeled as satisfying $$mu=mu_0+RTlnleft(frac{p_v}{p_0}right),$$

where $mu$ is the chemical potential (i.e., the molar Gibbs free energy, with $dmu=-s,dT+v,dP$ for a closed system, where $s$ is the molar entropy, $T$ is the temperature, $v$ is the molar volume, and $P$ is the system pressure), $mu_0$ is a reference value, $R$ is the gas constant, $p_v$ is the equilibrium vapor pressure in atm, and $p_0=1,mathrm{atm}$.

Differentiating at constant temperature, we obtain

$$dmu=frac{RT}{p_v}dp_v=v,dP;$$

$$frac{dp_v}{dP}=frac{vp_v}{RT}.$$

Therefore, for small changes in $p_v$, $$Delta p_vapproxfrac{vp_v}{RT}Delta P.$$

But $frac{vp_v}{RT}$ is generally a very small number ($10^{-8}$ for water at room temperature, for example). Thus, it may seem that the vapor pressure is not affected by the surrounding pressure—but it is.