This is a concrete version of Keith Kearnes's answer:

Let $A=f(X)$ be the image of a lattice monomorphism $f:X\to Y$ and consider in $Y\times\{0,1\}$ the partial order defined by $(x,i)\le (y,i)$ if $x\le y$ and for different $i,j\in\{0,1\}$ take $(x,i)\le (y,j)$ if there is $a\in A$ with $x\le a$ and $a\le y$. Now, consider the quotient $Z$ of $Y\times\{0,1\}$ which identifies points $(a,0)$ and $(a,1)$ for $a\in A$. Let $q$ be the corresponding quotient map $q:Y\times\{0,1\}\to Z$ and $g_i:Y\to Z$, $y\mapsto q(y,i)$.

$Z$ is a partially ordered set (maybe not a lattice because there are new incomparable $q(y,0)$ and $q(y,1)$ for $y\in Y\setminus A$) and $g_i$ preserve finite meets and joins: To see that, e.g., $q(x\vee y,0)=q(x,0) \vee q(y,0)$, one has to consider new upper bounds $q(z,1)$ of $\{q(x,0),q(y,0)\}$ with $z\notin A$. By definition of the partial order, there are $a,b\in A$ with $x\le a\le z$ and $y\le b\le z$ so that $A\ni a\vee b\le z$ which implies $q(x\vee y,0)\le q(a\vee b,0)=q(a\vee b,1)\le q(z,1)$.

Finally, let $i$ be the order embedding of $Z$ into its Dedekind-MacNeill completion $\hat Z$ which is a lattice (even a complete one). The embedding $i$ preserves meets and joins which exist in $Z$ and therefore, $i\circ g_j:Y\to \hat Z$ are lattice morphisms with $$\{y\in Y:i\circ g_0(y)=i\circ g_1(y)\}=\{y\in Y:g_0(y)=g_1(y)\}=A.$$ This implies that $f$ is a regular monomorphism.

This is a concrete version of Keith Kearnes's answer:

Let $A=f(X)$ be the image of a lattice monomorphism $f:X\to Y$ and consider in $Y\times\{0,1\}$ the partial order defined by $(x,i)\le (y,i)$ if $x\le y$ and for different $i,j\in\{0,1\}$ take $(x,i)\le (y,j)$ if there is $a\in A$ with $x\le a$ and $a\le y$. Now, consider the quotient $Z$ of $Y\times\{0,1\}$ which identifies points $(a,0)$ and $(a,1)$ for $a\in A$. Let $q$ be the corresponding quotient map $q:Y\times\{0,1\}\to Z$ and $g_i:Y\to Z$, $y\mapsto q(y,i)$.

$Z$ is a partially ordered set (maybe not a lattice because there are new incomparable $q(y,0)$ and $q(y,1)$ for $y\in Y\setminus A$) and $g_i$ preserve finite meets and joins: To see that, e.g., $q(x\vee y,0)=q(x,0) \vee q(y,0)$, one has to consider new upper bounds $q(z,1)$ of $\{q(x,0),q(y,0)\}$ with $z\notin A$. By definition of the partial order, there are $a,b\in A$ with $x\le a\le z$ and $y\le b\le z$ so that $A\ni a\vee b\le z$ which implies $q(x\vee y,0)\le q(a\vee b,0)=q(a\vee b,1)\le q(z,1)$.

Finally, let $i$ be the order embedding of $Z$ into its Dedekind-MacNeill completion $\hat Z$ which is a lattice (even a complete one). The embedding $i$ preserves meets and joins which exist in $Z$ and therefore, $i\circ g_j:Y\to \hat Z$ are lattice morphisms with $$\{y\in Y:i\circ g_0(y)=i\circ g_1(y)\}=\{y\in Y:g_0(y)=g_1(y)\}=A.$$ This implies that $f$ is a regular monomorphism.

Edit. Essentially the same arguments show that also in the category of complete lattices and meet and join preserving maps, all monomorphisms are regular.

This is a concrete version of Keith Kearnes's answer:

Let $A=f(X)$ be the image of a lattice monomorphism $f:X\to Y$ and consider in $Y\times\{0,1\}$ the partial order defined by $(x,i)\le (y_i)$ if $x\le y$ and for different $i,j\in\{0,1\}$ take $(x,i)\le (y,j)$ if there is $a\in A$ with $x\le a$ and $a\le y$. Now, consider the quotient $Z$ of $Y\times\{0,1\}$ which identifies points $(a,0)$ and $(a,1)$ for $a\in A$. Let $q$ be the corresponding quotient map $q:Y\times\{0,1\}\to Z$ and $g_i:Y\to Z$, $y\mapsto q(y,i)$.

$Z$ is a partially ordered set (maybe not a lattice because there are new incomparable $(y,0)$ and $(y,1)$ for $y\in Y\setminus A$) and $g_i$ preserve finite meets and joins: To see that, e.g., $q(x\vee y,0)=q(x,0) \vee q(y,0)$, one has to consider new upper bounds $q(z,1)$ of $\{q(x,0),q(y,0)\}$ with $z\notin A$. By definition of the partial order, there are $a,b\in A$ with $x\le a\le z$ and $y\le b\le z$ so that $A\ni a\vee b\le z$ which implies $q(x\vee y,0)\le q(a\vee b,0)=q(a\vee b,1)\le q(c,1)$.

Finally, let $i$ be the order embedding of $Z$ into its Dedekind-MacNeill completion $\hat Z$ which is lattice (even a complete one). The embedding $i$ preserves meets and joins which exist in $Z$ and therefore, $i\circ g_j:Y\to \hat Z$ are lattice morphisms with $$\{y\in Y:i\circ g_0(y)=i\circ g_1(y)\}=\{y\in Y:g_0(y)=g_1(y)\}=A.$$ This implies that $f$ is a regular monomorphism.

This is a concrete version of Keith Kearnes's answer:

Let $A=f(X)$ be the image of a lattice monomorphism $f:X\to Y$ and consider in $Y\times\{0,1\}$ the partial order defined by $(x,i)\le (y,i)$ if $x\le y$ and for different $i,j\in\{0,1\}$ take $(x,i)\le (y,j)$ if there is $a\in A$ with $x\le a$ and $a\le y$. Now, consider the quotient $Z$ of $Y\times\{0,1\}$ which identifies points $(a,0)$ and $(a,1)$ for $a\in A$. Let $q$ be the corresponding quotient map $q:Y\times\{0,1\}\to Z$ and $g_i:Y\to Z$, $y\mapsto q(y,i)$.

$Z$ is a partially ordered set (maybe not a lattice because there are new incomparable $q(y,0)$ and $q(y,1)$ for $y\in Y\setminus A$) and $g_i$ preserve finite meets and joins: To see that, e.g., $q(x\vee y,0)=q(x,0) \vee q(y,0)$, one has to consider new upper bounds $q(z,1)$ of $\{q(x,0),q(y,0)\}$ with $z\notin A$. By definition of the partial order, there are $a,b\in A$ with $x\le a\le z$ and $y\le b\le z$ so that $A\ni a\vee b\le z$ which implies $q(x\vee y,0)\le q(a\vee b,0)=q(a\vee b,1)\le q(z,1)$.

Finally, let $i$ be the order embedding of $Z$ into its Dedekind-MacNeill completion $\hat Z$ which is a lattice (even a complete one). The embedding $i$ preserves meets and joins which exist in $Z$ and therefore, $i\circ g_j:Y\to \hat Z$ are lattice morphisms with $$\{y\in Y:i\circ g_0(y)=i\circ g_1(y)\}=\{y\in Y:g_0(y)=g_1(y)\}=A.$$ This implies that $f$ is a regular monomorphism.