Consider the following problem:
Let $V$ be a $\mathbb{K}$-vector space. Let $v_1, v_2 \in V$ and $\lambda \in \mathbb{K}$. Show that $\langle v_1, v_2\rangle = \langle v_1, v_2 + \lambda v_1\rangle$, where $\langle S\rangle$ denotes the linear span of $S$.
It seems straigthforward to prove this in the following manner:
Let $S = \{v_1, v_2 \}, S' = \{v_1, v_2 + \lambda v_1 \}$. The span of $S'$ is
\begin{align*} \{x_1 v_1 + x_2(v_2 + \lambda v_1) \mid x_i\in \mathbb{K}\} &= \{x_1 v_1 + x_2 v_2 + \lambda x_1 \mid x_i\in \mathbb{K}\} \\ &= \{(x_1 + \lambda x_2)v_1 + x_2v_2 \mid x_i \in \mathbb{K}\} \end{align*}
Since $\delta := x_1 + \lambda x_2 \in \mathbb{K}$, $\{(x_1 + \lambda x_2)v_1 + x_2v_2\} = \{\delta v_1 + x_2 v_2\}$ is by definition the linear span of $S$ $\blacksquare$.
Once this proof was concluded, I attempted (for the sake of practicing and learning) to prove the spans are the same via the definition
The span of a set $S$ of vectors in a vector space $V$ is the interesection of all subspaces $W_1, ..., W_n$ of $V$ that contain $S$
In other words, I attempted to do the excercise thinking of the two spans as intersections instead of linear combinations. However, I haven't found a way to do it. How can one apply this "set-oriented" definition of linear span to prove the linear spans above are equal?
Translating your argument into this form is totally possible, but the underlying reasoning is very much the same.
Let $V$ be any subspace containing $S_1 = \{ v_1, v_2\}$. Since $V$ is a subspace (hence closed under linear combinations) we also have $v_2 + \lambda v_1 \in V$. So the subspace $V$ also contains $S_2 = \{v_1, v_2 + \lambda v_1 \}$.
Okay, what have we shown? We have shown that every subspace containing $S_1$ also contains $S_2$. In particular, the intersection of all of these, i.e. the span of $S_1$, contains $S_2$. This implies that it contains the span of $S_2$, i.e. $$ \langle S_1 \rangle \supseteq \langle S_2 \rangle. $$
Now let $V$ any subspace containing $S_2$, and reverse the argument above to show the opposite containment.