I am requried to prove that the following set of linear transformations is a basis for $\mathcal{L}(V,W)$ the set of all linear transformations from $V$ to $W$. I have established linear independence i am trying to prove that the given transformations span $\mathcal{L}(V,W)$ is my attempt correct?
Proposition. Let $V$ and $W$ be finite-dimensional vector spaces with ordered bases $\beta = \{v_1,v_2,\dots,v_n\}$ and $\gamma =\{w_1,w_2,\dots,w_m\}$, respectively. Now consider the transformations $T_{ij}:V\to W$ such that $$T_{ij}(v_k) = \begin{cases}w_i\text{ if }k=j\\\phantom{-}0\text{ if } k\neq j\end{cases}$$
Proof. Let $T\in\mathcal{L}(V,W)$, then since $\gamma$ is a basis for $W$ we may define for each $Tv_k = \sum_{i=1}^{m}\alpha_{ik}w_i$ for each $k\in\{1,2,\dots,n\}$. Now define the transformation $M = \left(\sum_{i=1}^{m}\sum_{j=1}^{n}\alpha_{ij}T_{ij}\right)$ then $Mv_k =\left(\sum_{i=1}^{m}\sum_{j=1}^{n}\alpha_{ij}T_{ij}\right)v_k = \sum_{i=1}^{m}\sum_{j=1}^{n}\alpha_{ij}T_{ij}v_k = \sum_{i=1}^{m}\alpha_{ik}T_{ik}v_k = \sum_{i=1}^{m}\alpha_{ik}w_i = Tv_k$ consequently by theorem $\textbf{2.6}$ it follows that $T=M$ and since our choice of $T$ was arbitrary it follows that $\operatorname{span}(T_{ij}) = W$.
Note: Theorem $2.6$ is the result that a linear transformation is uniquely determined by its action on the basis of a vector space.
Yes, it's fine.
A note about notation: I would write $\operatorname{span}\{T_{ij}\,|\,1\leqslant i\leqslant m,1\leqslant j\leqslant n\}$ instead of $\operatorname{span}T_{ij}$.