See Herbert Enderton, A Mathematical Introduction to Logic (2nd - 2001), page 23 :
take the special case in which $\Sigma$ is the empty set $\emptyset$.
Observe that it is vacuously true that any truth assignment satisfies every member of $\emptyset$. (How could this fail? Only if there was some unsatisfied member of $\emptyset$, which is absurd.) Hence we are left with : $\emptyset \vDash \psi$ iff every truth assignment (for the sentence symbols in $\psi$) satisfies $\psi$.
In this case we say that $\psi$ is a tautology (written $\vDash \psi$).
As highlighted by Enderton, we have a case of vacuous truth.
The condition for an interpretation $I$ to be a model of $\Sigma$ is that all the sentences in $\Sigma$ must be satisfied by the interpretation $I$.
This is :
for all $\sigma$, if $\sigma \in \Sigma$, then $I$ satisfy $\sigma$.
Thus, vacuous truth applies : there are no $\sigma \in \emptyset$.
What we are trying to convince ourselves is that : if $\emptyset \vDash \psi$, then $\psi$ is a tautology.
Consider again the definition of logically implies; there is a double conditional in place.
Saying that $\Sigma \vDash \psi$ means :
for every interpretation $I$, [ if for every sentence $\sigma$, ( if $\sigma \in \Sigma$, then $I$ satisfy $\sigma$ ), then $I$ satisfy $\psi$ ].
In semi-formal way :
$\forall I [ \forall \sigma(\sigma \in \Sigma \rightarrow I \vDash \sigma) \rightarrow I \vDash \psi ]$.
When we put $\emptyset$ in place of $\Sigma$, the antecedent of the "inner" conditional is false; thus, by truth-table for $\rightarrow$, the conditional is true (and this says nothing about the truth-value of the consequent !).
In this way, the antecedent of the "outer" conditional is true. But we are asserting the fact that $\emptyset \vDash \psi$, i.e. that the "outer" conditional is true.
If it is true and if its antecedent is true, there is only one possibility left : the consequent is true.
I.e.
$I \vDash \psi$.
This hold for every $I$, and thus we can conclude that $\psi$ is a tautology.
New addition
We can try with another approach.
According to the definition of logical consequence, $\varphi \vdash \psi$ iff $\varphi \land \lnot \psi$ is always false.
Now, assuming that $\Gamma$ is a finite set of sentences, i.e. $\Gamma = \{ \gamma_1, \ldots, \gamma_n \}$, we have that :
$\Gamma \vdash \psi$ iff $\gamma_1 \land \ldots \land \gamma_n \land \lnot \psi$ is always false.
Thus, if $\Gamma = \emptyset$, the above condition boils down to :
$\emptyset \vDash \psi$ iff $\lnot \psi$ is always false.
Obviously, $\lnot \psi$ is always false iff $\psi$ is a tautology.
Best Answer
No, your $U$ is not closed under logical consequence. For example, the statement $p$ is a logical consequence of $U$, but it is not in $U$. Same for $p \land p$, $p \land p \land p$, $z \lor \neg z$, etc.
Indeed, any set closed under logical consequence has to be of infinite size, so given that this set is of finite size tells you immediately that it is not closed under logical consequence.
Finally: you are trying to make some connection between satisfiability and logical consequence. I am not sure what connection you see, but there is this: if a set is not satisfiable, then every statement is a logical consequence of that set, and hence every statement should be in the logical closure of that set. In other words, a set is not satisfiable and closed under logical consequence if and only if it is the set of all logic statements.