Here is another way to construct such examples. This way one can get examples with Picard number at least up to $4$.

Notation: Let $S$ be a smooth projective surface, $\overline{NE}(S)$ the closed cone of effective curves, $Q_{\text{tot}}(S)$ be the set of numerical classes with positive self-intersection, and $Q(S)$ the closure of the connected component of $Q_{\text{tot}}(S)$ containing an ample class.

Fact (a simple consequence of Riemann-Roch): $Q(S)\subseteq \overline{NE}(S)$.

Claim Let $S$ be a smooth projective surface such that every proper curve $C$ on $S$ has postive self-intersection: $C^2>0$. Then for any two proper effective curves $C_1,C_2\subset S$ we have $C_1\cdot C_2>0$ and in particular $C_1\cap C_2\neq\emptyset$.

Remark: If the Picard number is not $1$, then the condition that "every proper curve $C$ on $S$ has postive self-intersection: $C^2>0$" is equivalent to assuming that $Q(S)=\overline{NE}(S)$ and that the boundary of $Q(S)$ is empty.

Proof By the nature of the statement we may assume that the Picard number of $S$ is at least $2$ and that $C_1$ and $C_2$ are irreducible. By the assumption $C_1^2>0$ and if it is irreducible, then this implies that it is nef. Hence the linear functional induced by $C_1\cdot(\quad)$ is non-negative on $Q(S)$, but since the boundary of $Q(S)$ is empty, it follows that $C_1\cdot(\quad)$ is actually positive on every effective class and hence the statement follows.

To see that there exists surfaces satisfying the condition in the Claim simply consider K3 surfaces that do not contain smooth rational or elliptic curves. These exist with Picard number 1-4. I suspect that one can find examples satisfying the condition in the Claim with higher Picard numbers, too.

Here is another way to construct such examples. This way one can get examples with Picard number at least up to $4$.

Notation: Let $S$ be a smooth projective surface, $\overline{NE}(S)$ the closed cone of effective curves, $Q_{\text{tot}}(S)$ be the set of numerical classes with positive self-intersection, and $Q(S)$ the closure of the connected component of $Q_{\text{tot}}(S)$ containing an ample class.

Fact (a simple consequence of Riemann-Roch): $Q(S)\subseteq \overline{NE}(S)$.

Claim Let $S$ be a smooth projective surface such that every proper curve $C$ on $S$ has postive self-intersection: $C^2>0$. Then for any two proper effective curves $C_1,C_2\subset S$ we have $C_1\cdot C_2>0$ and in particular $C_1\cap C_2\neq\emptyset$.

Remark: If the Picard number is not $1$, then the condition that "every proper curve $C$ on $S$ has postive self-intersection: $C^2>0$" is equivalent to assuming that $Q(S)=\overline{NE}(S)$ and that the boundary of $Q(S)$ does not contain any effective classes.

Proof By the nature of the statement we may assume that the Picard number of $S$ is at least $2$ and that $C_1$ and $C_2$ are irreducible. By the assumption $C_1^2>0$ and if it is irreducible, then this implies that it is nef. Hence the linear functional induced by $C_1\cdot(\quad)$ is non-negative on $Q(S)$, but since the boundary of $Q(S)$ does not contain any effective classes, it follows that $C_1\cdot(\quad)$ is actually positive on every effective class and hence the statement follows.

To see that there exists surfaces satisfying the condition in the Claim simply consider K3 surfaces that do not contain smooth rational or elliptic curves. These exist with Picard number 1-4. I suspect that one can find examples satisfying the condition in the Claim with higher Picard numbers, too.