Completeness of a space of continuous linear maps (TVS) 
Here are this $(\mathfrak{S}_1)$ and $(\mathfrak{S}_2)$ conditions:


*

*If $G_{1},G_{2}\in {\mathfrak {S}}$ then there exists a $G\in {\mathfrak {S}}$ such that $G_{1}\cup G_{2}\subseteq G$.

*If $G_{1}\in {\mathfrak {S}}$ and $\lambda$ is a scalar then there exists a $G\in {\mathfrak {S}}$ such that $\lambda G_{1}\subseteq G$.
I don't understand, why $E$ and $F$ have to be Hausdorff spaces. Where in this proof do we use their Hausdorff properties? And why do we need $E$ to be locally convex? 
 A: I think you need the range space $F$ Hausdorff to have unicity of the limits. Otherwise, in the first step of the proof, you would have to choose any of the many limits of $\mathscr F(x)$ and it is not clear that you can do this in a way to have the pointwise limit function $u$ linear.
I don't think that local convexity is essential for the theorem. It comes into play whenever you want to construct continuous linear operators via Hahn-Banach. 
A: Indeed neither local convexity nor Hausdorffness are needed in the theorem [assuming choice]. They are probably just assumed because that's the situation one is usually interested in.
The only condition of these that is used in the proof is that $F$ be Hausdorff, for that implies the uniqueness of the pointwise limit function $u$. If $F$ weren't Hausdorff there would be many pointwise limit functions, and one would need an explicit appeal to choice to show that there is at least one linear map among the pointwise limits.
So the proof as written works under the sole condition that $F$ be Hausdorff, provided the proof that $L_{\mathfrak{S}}(E;F)$ is a topological vector space didn't assume Hausdorffness or local convexity. [The fact doesn't need any such assumption, the proof given in the book might.]
To get rid of that condition, one must exhibit a linear limit function of $\mathscr{F}$. To that end, let $\mathscr{B}$ a basis (in the algebraic sense) of $E$ (first application of choice), and for every $b \in \mathscr{B}$ choose an $u_b$ such that $\mathscr{F}(b) \to u_b$ (second application of choice). Then define $u \colon E \to F$ as the linear map with $u(b) = u_b$ for every $b\in \mathscr{B}$ and verify that indeed $\mathscr{F}(x) \to u(x)$ for all $x\in E$.
