I can see this as clearly true because - assuming the action free - if $g_1 \not=g_2$, then $g_1 \cdot x \not= g_2 \cdot x$ for all $x \in X$ implies it holds true for some $x \in X$, so the actions is faithful.
However, our teacher defined faithful actions interms of the action induced homomorphism $ \phi : G \to S(X)$ and called the action faithful whenever $\phi$ is injective.
To prove the map injective, how should I proceed? If I define $\phi(g)=g \cdot x$, then $$g_1 \not= g_2 \implies \phi(g_1)\not= \phi(g_2)$$ proves $\phi$ is injective. I'm not sure if the mapping is defined correctly, however.
Also can someone show via an example how a faithful action isn't always free?