Proving a linear transformation is unique

By the way, the proof given above does indeed show that $T$ is unique. Perhaps a different phrasing of the argument might help you understand the uniqueness.

You said you understood the existence part, so ok let $T: V \to W$ be the linear map you constructed, which satisfies $T(v_j) = w_j$ for all $j \in \{1, \dots, n\}$. Suppose now that there is a linear map $S: V \to W$ such that $S(v_j) = w_j$ for all $j \in \{1, \dots, n\}$. We have to show that $T=S$; i.e we have to show that for every $x \in V$, $T(x) = S(x)$.

To prove this, pick any $x \in V$. Since $\{v_1, \dots, v_n\}$ is a basis for $V$, there exist (unique) scalars $c_1, \dots, c_n \in F$ such that \begin{align} x = \sum_{i=1}^n c_iv_i \tag{*} \end{align} Now, we perform a simple computation: \begin{align} T(x) &= T\left( \sum_{i=1}^n c_iv_i \right) \tag{by (*)} \\ &= \sum_{i=1}^n c_i T(v_i) \tag{$T$ is linear by construction} \\ &= \sum_{i=1}^n c_i w_i \tag{by definition of $T$} \\ &= \sum_{i=1}^n c_i S(v_i) \tag{by assumption on $S$} \\ &= S\left( \sum_{i=1}^n c_i v_i \right) \tag{$S$ linear by assumption} \\ &= S(x) \tag{by (*)} \end{align} So we have shown that for every $x \in V$, $T(x) = S(x)$. Hence, $T=S$, proving uniqueness of the original $T$ you constructed.

This result is often stated as "a linear transformation is specified by its values on a basis" or something to that effect.