Is "The empty set is a subset of any set" a convention?

Recently I learned that for any set A, we have $\varnothing\subset A$.

I found some explanation of why it holds.

$\varnothing\subset A$ means "for every object $x$, if $x$ belongs to the empty set, then $x$ also belongs to the set A". This is a vacuous truth, because the antecedent ($x$ belongs to the empty set) could never be true, so the conclusion always holds ($x$ also belongs to the set A). So $\varnothing\subset A$ holds.

What confused me was that, the following expression was also a vacuous truth.

For every object $x$, if $x$ belongs to the empty set, then $x$ doesn't belong to the set A.

According to the definition of the vacuous truth, the conclusion ($x$ doesn't belong to the set A) holds, so $\varnothing\not\subset A$ would be true, too.

Which one is correct? Or is it just a convention to let $\varnothing\subset A$?

There’s no conflict: you’ve misinterpreted the second highlighted statement. What it actually says is that $\varnothing$ and $A$ have no element in common, i.e., that $\varnothing\cap A=\varnothing$. This is not the same as saying that $\varnothing$ is not a subset of $A$, so it does not conflict with the fact that $\varnothing\subseteq A$.

To expand on that a little, the statement $B\nsubseteq A$ does not say that if $x\in B$, then $x\notin A$; it says that there is at least one $x\in B$ that is not in $A$. This is certainly not true if $B=\varnothing$.

From Halmos's Naive Set Theory:

A transcription:

The empty set is a subset of every set, or, in other words, $\emptyset \subset A$ for every $A$. To establish this, we might argue as follows. It is to be proved that every element in $\emptyset$ belongs to $A$; since there are no elements in $\emptyset$, the condition is automatically fulfilled. The reasoning is correct but perhaps unsatisfying. Since it is a typical example of a frequent phenomenon, a condition holding in the "vacuous" sense, a word of advice to the inexperienced reader might be in order. To prove that something is true about the empty set, prove that it cannot be false. How, for instance, could it be false that $\emptyset \subset A$? It could be false only if $\emptyset$ had an element that did not belong to $A$. Since $\emptyset$ has no elements at all, this is absurd. Conclusion: $\emptyset \subset A$ is not false, and therefore $\emptyset \subset A$ for every $A$.

What confused me was that, the following expression was also a vacuous truth.

For every object of $x$, if $x$ belongs to the empty set, then $x$ doesn't belong to the set $A$.

As a complement (heh) to Brian Scott's (+1) answer, your argument shows that $\varnothing \subset A^{c}$, the complement of $A$. This statement is also (vacuously) true.

Very subtle point:

"All x are not something" does not imply "Not all x are something".

The first may be vacuously true. The second one can not. If the x are vacuous then the second one has to be "vacuously false" as all x of nothing are any property so it is impossible for them not to be any property.

So "all elements of the empty set are not in $S$" does not imply "Not all elements of the empty set are in $S$" $\iff$ "It is not true that all elements of the empty set are in $S$" $\iff$ "There are some elements of the empty set that are not in $S$".

The first is vacuously true (and is equivalent to $\emptyset \subset S^c$ which is true) and the second set of equivalent statements are all equivalent to $\emptyset \not \subset S$ which is not true.

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The thing is what you say is absolutely correct for non empty sets.

More formally:

All elements $x$ in $S$ are not in $A$ $\implies$

$S \subset A^c$ $\implies$

$\color{red}{\text{There is an } x\in S \text{ where } x \not \in A}\implies$

It is not true that all $x \in S$ are also in $A$ $\implies$

$S \not \subset A$.

However the red line can only be concluded if $A$ is non-empty. If $A$ is empty the red line is simply false.

And without the red line there is simply no logic or means to jump from the line before to the line after:

All elements $x$ in $S$ are not in $A$ $\not\implies$

It is not true that all $x \in S$ are also in $A$.

That simply is not true for an empty $S$.