Weak (initial) topology induced by a family of functions

Let $S$ be a set and let $F$ be a family of real valued functions on $S$ such that $f(s_{1}) = f(s_{2})$ for all $f \in F$ implies $s_{1} = s_{2}$. Prove that there exists a weakest topology in $S$ amongst all those for which all members of $F$ are continuous. Show further that the resulting topological space satisfies the Hausdorff separation axiom.

Prove Proposition 12.

Proposition 12: Let $X$ be a non-empty set and $\mathcal{F} = \left\{ f_{\lambda }: X \rightarrow X_{\lambda }\right\}_{\lambda \in \Lambda }$ be a collection of mappings parametrized by a set $\Lambda$, where each $X_{\lambda }$ is a topological space. The topology $\mathcal{T}$ for $X$ that has the collection of sets $\mathcal{C} = \left\{ f_{\lambda }^{-1}(O_{\lambda }) \mid \lambda \in \Lambda , O_{\lambda } \text{ open in } X_{\lambda }\right\}$ as a subbase is the weakest among the topologies on $X$ for which each mapping $f_{\lambda }: X \rightarrow X_{\lambda }$ is continuous. We call $\mathcal{T}$ the weak topology induced by $\mathcal{F}$.

Let $\mathcal{F}$ be a collection of real-valued functions on a set $X$. Find necessary and sufficient conditions on $\mathcal{F}$ in order that $X$, considered as a topological space with the weak topology induced by $\mathcal{F}$, is Tychonoff.

Let ($X, \mathcal{T}$) be a normal topological space and $\mathcal{F}$ the collection of continuous real-valued functions on $X$. Show that $\mathcal{T}$ is the weak topology induced by $\mathcal{F}$.

Suppose that $X$ is a topological space for which there is a collection of continuous real-valued functions on $X$ that separates points in $X$. Show that $X$ is Hausdorff.