Here's a point of view that combines aspects of both of the current (good) answers but makes things more explicit.
Take $U$ an affine open piece of $X$ containing $x$, $U \cong \operatorname{Spec}(A)$, with $x$ corresponding to $P < A$. Then the localisation map $\phi:A\rightarrow A_P$ induces a morphism $\operatorname{Spec}(\mathcal{O}_{X,x})\rightarrow \operatorname{Spec}(A)\rightarrow X$. The image of this morphism is the set of primes in $A$ that are themselves contained in $P$. (Using properties of localisation of rings of morphisms between affine schemes.) This is equal to the intersection of every open subset of $\operatorname{Spec}(A)$ containing $P$, as is easily verified by checking for distinguished affine open subsets of $U$, $D(f)$ for $f \in A\setminus P$.
Now suppose $f \in A/ P$. Then the image of $\phi$ lies inside the distinguished open affine piece $D(f)$. Equivalently, $\phi$ factors through $A_f$, i.e. there is $\phi_f:A_f\rightarrow A_P$, such that if $i_f:A\rightarrow A_f$ is the localisation map, $\phi = \phi_f\circ i_f$. This means that the morphism $\operatorname{Spec}(\mathcal{O}_{X,x})\rightarrow\operatorname{Spec}(A_f)\rightarrow X$ induced by taking affine open piece $U' \cong \operatorname{Spec}(A_f)$ of $X$ is equal to the morphism induced by $U$. (Here we are implicitly using that for a manipulatively closed subset $S$ of a ring $R$, with $P$ a prime not meeting $S$, $(S^{-1}R)_{S^{-1}P}$ is naturally isomorphic to $R_P$.)
Thus if $V \cong \operatorname{Spec}(B)$ is another open affine piece of $X$ containing $x$, then taking open $W \subset U\cap V$ a distinguished open affine piece of both $U$ and $V$ (i.e. there is $a \in A, b \in B$ such that $W \cong D(a)$ and $W \cong D(b)$), the morphism $\operatorname{Spec}(\mathcal{O}_{X,x})\rightarrow X$ induced by $A$ is equal to the morphism induced by $A_a$ by the above paragraph, which is equal to the morphism induced by $B_b$ from the construction of $W$, which is then equal to the morphism induced by $B$ using the above paragraph again.
So we have shown that the morphism is independent of choice of open affine piece. Given this, the image is immediately seen to be the intersection of every open set about $x$, and (interestingly) this view is completely independent of the choice of affine piece.