Since MathJax does not seem to support \bigsqcap anyway, I’ll write meets using the standard $\bigwedge$ notation.

Logic compactness is essentially equivalent to well foundedness:

Proposition. Let $L$ be a lattice.

1. If $L$ is well founded, then for every nonempty $\Gamma\subseteq L$, $\bigwedge\Gamma$ exists, and equals $\bigwedge\Gamma_0$ for some finite $\Gamma_0\subseteq\Gamma$. Thus, $L$ is logic-compact.
2. If $\bigwedge\Gamma$ exists for every nonempty $\Gamma\subseteq L$, and $L$ is logic-compact, then $L$ is well founded.
3. If $\bigwedge\Gamma$ does not necessarily exist, then I do not know how to interpret the definition of logic compactness in the first place.

Proof:

1. $\{\bigwedge\Gamma_0:\varnothing\ne\Gamma_0\subseteq\Gamma\text{ finite}\}\subseteq L$ is nonempty, hence it has a minimal element $\bigwedge\Gamma_0$ by well foundedness. Then $\bigwedge\Gamma_0$ is $\bigwedge\Gamma$, i.e., $\bigwedge\Gamma_0\le x$ for each $x\in\Gamma$, lest $\bigwedge(\Gamma_0\cup\{x\})<\bigwedge\Gamma_0$, contradicting minimality.

2. If $L$ is not well founded, there is a strictly decreasing chain $x_0>x_1>x_2>\cdots$. Put $\Gamma=\{x_n:n\in\omega\}$ and $y=\bigwedge\Gamma$. Then $y<\bigwedge\Gamma_0=\min\Gamma_0\in\Gamma$ for any nonempty finite $\Gamma_0\subseteq\Gamma$, witnessing that $L$ is not logic-compact.

3. By fiat.

While the question is still unclear, it seems now that it is very different from the question answered below, which was based on my best guess from reading the first revision.

Since MathJax does not seem to support \bigsqcap anyway, I’ll write meets using the standard $\bigwedge$ notation.

Logic compactness is essentially equivalent to well foundedness:

Proposition. Let $L$ be a lattice.

1. If $L$ is well founded, then for every nonempty $\Gamma\subseteq L$, $\bigwedge\Gamma$ exists, and equals $\bigwedge\Gamma_0$ for some finite $\Gamma_0\subseteq\Gamma$. Thus, $L$ is logic-compact.
2. If $\bigwedge\Gamma$ exists for every nonempty $\Gamma\subseteq L$, and $L$ is logic-compact, then $L$ is well founded.
3. If $\bigwedge\Gamma$ does not necessarily exist, then I do not know how to interpret the definition of logic compactness in the first place.

Proof:

1. $\{\bigwedge\Gamma_0:\varnothing\ne\Gamma_0\subseteq\Gamma\text{ finite}\}\subseteq L$ is nonempty, hence it has a minimal element $\bigwedge\Gamma_0$ by well foundedness. Then $\bigwedge\Gamma_0$ is $\bigwedge\Gamma$, i.e., $\bigwedge\Gamma_0\le x$ for each $x\in\Gamma$, lest $\bigwedge(\Gamma_0\cup\{x\})<\bigwedge\Gamma_0$, contradicting minimality.

2. If $L$ is not well founded, there is a strictly decreasing chain $x_0>x_1>x_2>\cdots$. Put $\Gamma=\{x_n:n\in\omega\}$ and $y=\bigwedge\Gamma$. Then $y<\bigwedge\Gamma_0=\min\Gamma_0\in\Gamma$ for any nonempty finite $\Gamma_0\subseteq\Gamma$, witnessing that $L$ is not logic-compact.

3. By fiat.