It is well-known that a finite join-semi-lattice with a maximum element is a lattice.

But this is not true for infinite posets. Let $P := (\{(a,b)\colon 0\leq a,b \leq 1\}\setminus \{(1,1)\}) \cup \{(a,a)\colon 1 < a \leq 2\}$, with the usual partial order $(a_1,b_1)\leq (a_2,b_2)$ iff $a_1 \leq a_2$ and $b_1 \leq b_2$. Then $P$ is a join-semi-lattice (with $(a_1,b_1)\wedge (a_2,b_2)=(\mathrm{min}(a_1,a_2),\mathrm{min}(b_1,b_2)$) and it has a maximum element $(2,2)$. But $(1,0)$ and $(0,1)$ lack a least upper bound.

It is well-known that a finite meet-semi-lattice with a maximum element is a lattice. The reason is that we can define $a \vee b := \wedge \{c\colon \textrm{$c$ is an upper bound for $a,b$}\}$, where this set is non-empty (since we have a maximum) and finite (since the poset is finite), and finite meets exist by supposition that we have a meet-semi-lattice.

But this is not true for infinite posets. Let $P := (\{(a,b)\colon 0\leq a,b \leq 1\}\setminus \{(1,1)\}) \cup \{(a,a)\colon 1 < a \leq 2\}$, with the usual partial order $(a_1,b_1)\leq (a_2,b_2)$ iff $a_1 \leq a_2$ and $b_1 \leq b_2$. Then $P$ is a meet-semi-lattice (with $(a_1,b_1)\wedge (a_2,b_2)=(\mathrm{min}(a_1,a_2),\mathrm{min}(b_1,b_2)$) and it has a maximum element $(2,2)$. But $(1,0)$ and $(0,1)$ lack a join.