For basis $\mathcal{B}$ of the topology on $X$, the restriction functor $\mathbf{r}:{\rm Sh}(X)\to{\rm Sh}(\mathcal{B})$ is an equivalence. This is Exercise II.4 of Mac Lane and Moerdijk's, "Sheaves in Geometry and Logic [. . .]". According to Approach0, it is new to MSE.
The Details:
Adapted from p. 13, ibid. . . .

Definition 1: A functor $F:  \mathbf{A}\to \mathbf{B}$ is an equivalence of categories if for any $\mathbf{A}$-objects $A, A'$, we have that
$$\begin{align}
{\rm Hom}_{\mathbf{A}}(A, A')&\to{\rm Hom}_{\mathbf{B}}(FA, FA')\\
p&\mapsto F(p)
\end{align}$$
is a bijection and, moreover, any object of $\mathbf{B}$ is isomorphic to an object in the image of $F$.

On p. 66, ibid. . . .

Definition 2: A sheaf of sets $F$ on a topological space $X$ is a functor $F:\mathcal{O}(X)^{{\rm op}}\to\mathbf{Sets}$ such that each open covering $U=\bigcup_iU_i, i\in I$, of open subsets of $U$ of $X$ yields an equaliser diagram 
$$ FU\stackrel{e}{\dashrightarrow}\prod_{i\in I}FU_i\overset{p}{\underset{q}{\rightrightarrows}}\prod_{i,j\in I}(U_i\cap U_j),$$
where for $t\in FU,$ $e(t)=\{ t\rvert_{U_i}\mid i\in I\}$ and for a family $t_i\in FU_i$,
$$p\{ t_i\}=\{t_i\rvert_{(U_i\cap U_j)}\}\quad\text{ and }\quad q\{ t_i\}=\{t_j\rvert_{(U_i\cap U_j)}\}.$$

Here $\mathcal{O}(X)$ is the set of open sets of $X$.
The Question:

Prove that for a basis $\mathcal{B}$ of the topology on a space $X$, the restriction functor $\mathbf{r}:{\rm Sh}(X)\to{\rm Sh}(\mathcal{B})$ is an equivalence of categories.
[Hint: Define a quasi-inverse $\mathbf{s}:{\rm Sh}(\mathcal{B})\to{\rm Sh}(X)$ for $\mathbf{r}$ as follows. Given a sheaf $F$ on $\mathcal{B}$, and an open set $U\subset X$, consider the cover $\{B_i\mid i\in I\}$ of $U$ by all basic open sets $B_i\in\mathcal{B}$ which are contained in $U$. Define $\mathbf{s}(F)(U)$ by the equaliser 
$$\mathbf{s}(F)(U)\to\prod_{i\in I}F(B_i)\rightrightarrows\prod_{i, j}F(B_i\cap B_j).]$$

Thoughts:
I need to show that, following Definition 1,
$$\begin{align}
{\rm Hom}_{{\rm Sh}(X)}(V, V')&\to{\rm Hom}_{{\rm Sh}(\mathcal{B})}(\mathbf{r}V, \mathbf{r}V')\\
p&\mapsto \mathbf{r}(p)
\end{align}$$ is a bijection for all $V, V'\in{\rm Ob}({\rm Sh}(X))$ and any ${\rm Sh}(\mathcal{B})$-object is isomorphic to an object in the image of $\mathbf{r}$.
Further Context:
Related questions of mine include:


*

*Equivalence of categories preserves subobject classifiers.

*Equivalence of a CCC with another category means that that category is also a CCC.
I am teaching myself topos theory for fun. I have read Goldblatt's book, "Topoi [. . .]", although I did not fully understand it. For example, 


*

*Reading Chapter 14 of Goldblatt's, "Topoi: A Categorial Analysis of Logic."
Please help :)
 A: First of all, when the hint speaks of a "quasi-inverse" it is referring to the following equivalent of the given definition: a functor $F : \mathbf{C} \to \mathbf{D}$ is an equivalence of categories if and only if there exists a functor $G : \mathbf{D} \to \mathbf{C}$ such that $F \circ G \simeq \operatorname{id}_{\mathbf{D}}$ and $G \circ F \simeq \operatorname{id}_{\mathbf{C}}$; and in this case, $G$ is called a quasi-inverse of $F$.
So, one way to follow the hint would be to explain how $\mathbf{s}$ becomes a functor (i.e. how it operates on morphisms, and show it preserves identities and compositions), and then establish isomorphisms $\mathbf{r} \circ \mathbf{s} \simeq \operatorname{id}$ and $\mathbf{s} \circ \mathbf{r} \simeq \operatorname{id}$.

On the other hand, it is possible to proceed using the definition you stated.  First, as a preliminary, I don't know if MacLane and Moerdik specified what exactly $\operatorname{Sh}(\mathcal{B})$ means; but the reasonable definition would be that it is the presheaves on the poset category of $\mathcal{B}$ such that whenever $\{ V_i \mid i \in I \} \subseteq \mathcal{B}$ is a cover of $U \in \mathcal{B}$, we have an equalizer diagram
$$F(U) \rightarrow \prod_{i\in I} F(V_i) \rightrightarrows \prod_{i, j \in I, W\in \mathcal{B}, W \subseteq V_i \cap V_j} F(W).$$
(The first step would be to see why $\mathbf{r}$ of a sheaf on $X$ would satisfy this condition; I will leave that as an exercise.)
So, first let us see that $\mathbf{r}$ is injective on morphisms; so, suppose that we have two morphisms $f, g : F \to G$ such that $f(V) = g(V)$ whenever $V \in \mathcal{B}$.  Then for any open $U$ and $x \in F(U)$, there is a cover of $U$ by elements $\{ V_i \mid i \in I \} \subseteq \mathcal{B}$.  Now, by the hypothesis, $$f(x) {\mid_{V_i}} = f(V_i)(x {\mid_{V_i}}) = g(V_i)(x {\mid_{V_i}}) = g(x) {\mid_{V_i}}$$ for each $i$; and by the injectivity part of the equalizer condition defining that $G$ is a sheaf, we conclude that $f(x) = g(x)$.  Since this is true for any open $U$ and any $x \in F(U)$, then $f = g$.
Similarly, to see that $\mathbf{r}$ is surjective on morphisms, suppose we have $f : \mathbf{r}(F) \to \mathbf{r}(G)$.  Then for any open $U \subseteq X$ and $x \in F(U)$, again choose a cover of $U$ by $\{ V_i \mid i \in I \} \subseteq \mathcal{B}$.  (In fact, to forestall questions of the following construction being well-defined, let us use the canonical maximal cover of all elements of $\mathcal{B}$ contained in $U$.)  Then for each $i \in I$, define $y_i := f(V_i)(x {\mid_{V_i}})$.  Then for each $i,j$, we can find the canonical maximal cover of $V_i \cap V_j$ by $\{ W_k \mid k \in K_{i,j} \} \subseteq \mathcal{B}$.  Now for each $k$, we have
$$y_i {\mid_{W_k}} = f(V_i)(x {\mid_{V_i}}) {\mid_{W_k}} = f(W_k)((x {\mid_{V_i}}) {\mid_{W_k}}) = F(W_k)(x {\mid_{W_k}}) = y_j {\mid_{W_k}}.$$
Therefore, by the injectivity part of the sheaf condition on $G$, we have $y_i {\mid_{U_i \cap U_j}} = y_j {\mid_{U_i \cap U_j}}$.  Then, by the exactness part of the sheaf condition on $G$, there exists a unique $y \in G(U)$ such that $y {\mid_{U_i}} = y_i$.  We now define $f'(U)(x) := y$.
It remains to show that $f'$ defines a morphism of sheaves, and that $\mathbf{r}(f') = f$.  (Hint for the morphism of sheaves part: given $U' \subseteq U$ and $x \in F(U)$, show that $(f'(U) {\mid_{U'}}) {\mid_{V_i}}$ is equal to $y_i$ when you put $x {\mid_{U'}}$ in place of $x$, and then apply the injectivity part of the sheaf condition on $G$.)
Now, to show that $\mathbf{r}$ is essentially surjective, suppose we have $F \in \operatorname{Sh}(\mathcal{B})$.  Then for each open $U$, define $G(U)$ to be the equalizer in the diagram
$$G(U) \rightarrow \prod_{V \in \mathcal{B}, V \subseteq U} F(V) \rightrightarrows \prod_{V, V', W \in \mathcal{B}, V \subseteq U, V' \subseteq U, W \subseteq V \cap V'} F(W).$$
The restriction maps of $G$ will then be constructed based on the universal property of equalizers.  We now need to see that $G$ is a sheaf on $X$, and that $\mathbf{r}(G) \simeq F$.  The latter follows fairly directly from the sheaf condition on $F$.
For the sheaf condition, suppose we have a cover $\{ U_i \mid i \in I \}$ of $U$ and sections $x_i \in G(U_i)$ such that $x_i {\mid_{U_i \cap U_j}} = x_j {\mid_{U_i \cap U_j}}$ for each $i,j$.  Then each $x_i$ can be decomposed into the compatible data of an element of $F(V)$ for every $V \in \mathcal{B}$, $V \subseteq U_i$ which we will call $x_i {\mid_V}$.  But then, the union of the canonical covers of each $U_i$ will form a cover of $U$; and for each $W$ in this cover, we can choose $i$ such that $W \subseteq U_i$, and define $y_W := x_i {\mid_W}$.  If we have two different indices $i,j$ such that $W \subseteq U_i$ and $W \subseteq U_j$, then from the condition $x_i {\mid_{U_i \cap U_j}} = x_j {\mid_{U_i \cap U_j}}$ we get $x_i {\mid_W} = x_j {\mid_W}$, which makes this definition of $y_V$ well-defined.  Once we verify the compatibility condition on $(y_W)$, we get a section $z_V \in F(V)$ from the definition of $F$ being a sheaf.  It now remains to show that this family of $z_V$ satisfies the compatibility condition from the definition of $G$, and that the section $x \in G(U)$ we get in this way satisfies $x {\mid_{U_i}} = x_i$ for each $i$.  It also remains to establish the uniqueness of $x$.

In the above, you can see that our construction in the "essential surjectivity" proof amounted to specifying the object part of a quasi-inverse $\mathbf{s}$, and our construction in the "surjectivity on morphisms" proof amounted to specifying the morphism part of $\mathbf{s}$.  (Note that the definition of $\mathbf{s}$ as you wrote it does not necessarily make sense if $\mathcal{B}$ is not closed under intersections.)
