Does weak interaction always come in decays? Well, essentially I'm confused about how weak nuclear force works. Decay is usually mentioned in studying weak nuclear force, so does weak force always work in decays? If so, is all decay is based on the weak nuclear force?
 A: Some decays are caused by the electromagnetic force. For example, the neutral pion, $\pi^0$, made of a superposition of $u\bar{u}$ and $d\bar{d}$ decays by annihilation of the quark-antiquark into 2 photons. Other decays happen through the strong interaction. For example the neutral Delta baryon, $\Delta^0$, made of $udd$, i.e. the same make-up as the neutron, but the constituent are in a higher energy state than the neutron. The decay $\Delta^0 \to \pi^- + p$ happens through the strong interaction. I'll add the Feynman diagram later.
A note in passing: since the strong interaction is much stronger than the electromagnetic interaction which is much stronger than the weak interaction, the particle decaying through the strong interaction tend to have a shorter lifetime than those decaying through the electromagnetic interaction which tend to have a shorter lifetime than those decaying through the weak interaction.
As your first question, I guess your point is whether the weak interaction could make bound states for example. The answer is "no". The weak interaction only appears in decays indeed.
A: If by decay you mean the process whereby a particle at rest breaks apart into two or more constituents each of which carries energy and momentum, then yes, the weak interaction involves more than just decay events.  There are also weak scattering events such as neutrino elastic scattering (see here).  There is also electron capture whereby a proton bound to a nucleus captures an atomic electron (s-state) leaving a neutron and an outgoing neutrino. Admittedly this latter process also involves the strong interaction so it is not entirely a weak provess. Finally, there is inelastic neutrino scattering (also known as inverse beta-decay, see here)
