Span of a subset of a vector space is the smallest subspace containing that set
Theorem: Let $S$ be a subset of a vector space $V$. Then $\text{span}(S)$ is the smallest vector subspace of $V$ which contains $S$.
What kind of set $T$ would disprove the theorem? It would have to be (1) a vector subspace which (2) contains every vector in $S$, and which (3) is smaller than span(S) so that $T\subsetneq\text{span}(S)$.
Let's suppose $T$ has the first two qualifications. We'll show that it can't have the third qualification; that's enough to establish the proof. In particular, suppose:
$T$ is not just a set; it's a vector subspace. This means that $T$ is closed under linear combinations. (every linear combination of vectors in $T$ also belongs to $T$.)
$T$ contains every member of $S$. ($T \supseteq S$).
By the first two properties, $T$ contains every vector in $S$ and $T$ is closed under linear combinations, so it follows that $T$ contains every linear combination of vectors in $S$. But $\text{span}(S)$ is by definition the set of all linear combinations of vectors in $S$. Therefore $T\supseteq \text{span}(S)$. Therefore $T$ is at least as large as $\text{span}(S)$, so the theorem is true.
Here is your statement written slightly different. Maybe that will help.
Theorem: If $W\subset V$ is a subspace, such that $v_1,…,v_k∈W,$ then $\text{span}(v_1, …,v_k)\subset W$. Proof: Since $v_1,…,v_k∈W$ and $W$ is a subspace all linear combinations $$α_1v_1 + … + α_kv_k∈W$$ Since $\text{span}(v_1,…,v_k)$ contains (only) these linear combination it follows $\text{span}(v_1,…,v_k)\subset W$.
Written in words that theorem states, that any subspace $W$, that exists and contains $v_1,…,v_n$, also contains $\text{span}(v_1,…,v_k)$. Hence $W$ is larger (or equal) to $\text{span}(v_1,…,v_k)$, since it contains $\text{span}(v_1,…,v_k)$ and could contain some more elements.
And since any other subspace $W$ is larger (or equal) it follows that $\text{span}(v_1,…,v_k)$ the smallest subspace.
That is like saying: Any number in $ℕ∪\{0\}$ is larger (or equal) to $0$, hence $0$ is the smallest number in $ℕ∪\{0\}$.
Or even closer to the original problem: Any subset $M⊂(ℕ∪\{0\})$ with $0∈M$ is larger or equal to $\{0\}$, hence $\{0\}$ has to be the smallest subset of $ℕ∪\{0\}$ that contains $0$.