Suppose $I$ is the set of injective mappings $A \to B$.
To see $I$ is closed we apply the last criterion. So let $g$ be a map such that
$$\forall E \subseteq A \text{ finite }: \exists f \in I: f\restriction_E = g\restriction_E \tag{1}$$
and we need to show that $g \in I$, so $g$ is injective. Let $a_1 \neq a_2$ in $A$. Set $E= \{a_1,a_2\}$ and by $(1)$ we have some $f \in I$ such that $f\restriction_E = g\restriction_E$ and as $f$ is injective $g(a_1) = f(a_1) \neq f(a_2) = g(a_2)$ and so $g$ assumes different values on $a_1$ and $a_2$ too. As $a_1$ and $a_2$ were arbitrary $g$ is indeed injective, and we're done. $I$ is closed.
(you could go by contradiction too: assume $g(a)=g(a')$ and use $E=\{a,a'\}$ as before, but that's a matter of taste.)
That $B^A$ is Hausdorff is quite easy: suppose $f \neq g$ are two points in $B^A$. The fact that the functions are different means that there is some $a \in A$ such that $f(a) \neq g(a)$.
Define $$U(f) = \{ h \in B^A: h(a)=f(a)\} \text{ and } U(g)=\{h \in B^A: h(a)=g(a)\}$$
Then clearly $U(f) \cap U(g)= \emptyset$ and it suffices to show that $U(f)$ (and similarly $U(g)$) are open: we apply the first criterion and for each $h$ in $U(f)$ (or $U(g)$) we take $E=\{a\}$ (a finite subset of $A$), and by definition all $h' \in B^A$ with $h\restriction_E = h'\restriction_E$ have the same value as $h$ on the point $a$ and so also belong to that same $U(f)$ resp. $U(g)$ as $h$ did. This finishes the proof of Hausdorffness.
To treat $C:=\textrm{Aut}_K(L)$ (assuming $K \subseteq L$ is a field extension) we see $C$ as a subspace of $L^L$.
To see it is closed, we again use the third criterion: suppose $g$ is such that
$$\forall E \subseteq L \text{ finite }: \exists f \in C: f\restriction_E = g\restriction_E \tag{2}$$
and we need to show $g \in C$, i.e. $g$ is a $K$-automorphism of $L$, so
we need to check the field homomorphism conditions for $g$ from $L$ to $L$ and the fact that $g$ fixes $K$ pointwise. I'll do a few for you to get the spirit:
So we need to check $\forall l_1,l_2 \in L: g(l_1)+g(l_2)= g(l_1+l_2)$, so pick arbitrary $l_1,l_2 \in L$. Define $E = \{l_1,l_2,l_1+l_2\}$ and apply $(2)$ to get $f \in C$ such that $f$ and $g$ agree on $L$.
It follows that $g(l_1+l_2) = f(l_1 + l_2)=$ (as $l_1,+l_2 \in E$) $=f(l_1)+f(l_2) = $ (as $f$ is a $K$-automorphism) $= g(l_1) + g(l_2)$ (as $l_1,l_2 \in E$) and so the condition on $g$ is checked for these points. As we can do this for all $l_1,l_2 \in L$, $g$ preserves addition. Note that we only need finitely many points at the time and so the criterion for closedness can be succesfully applied.
The case for preservation of $\cdot$ is almost identical, whcih makes $g$ a group homomorphism. The condition $\forall k \in K: g(k)=k$ can be checked by using $E=\{k\}$ of course, for each arbitrary $k \in K$: if $f$ agrees with $g$ on $\{k\}$, $g(k)=f(k)=k$.
So by this criterion, $\textrm{Aut}_K(L)$ is closed in $L^L$.
