Is my proof of the following proposition correct?
Proposition. Given that $V_1,V_2,\dots,V_m$ are vector spaces such that $V_1\times V_2\times\cdots\times V_m$ is finite dimensional. Prove that $V_j$ is finite dimensional for each $j = 1,2,\dots,m$.
Proof. Let $(u_{11},u_{12},\dots,u_{1m}),(u_{21},u_{22},\dots,u_{2m}),\dots,(u_{m1},u_{m2},\dots,u_{mm})$ be the basis of $V_1\times V_2\times\cdots\times V_m$. Now let $j\in\{1,2,\dots,m\}$, we show that $V_j = \operatorname{span}(u_{1j},u_{2j},\dots,u_{mj})$.
Assume that $w\in V_{j}$, consequently $(0,0,\dots,w_j,\dots,0)\in V_1\times V_2\times\cdots\times V_m$ and thus for some $\lambda_1,\lambda_2,\dots,\lambda_m\in\mathbf{F}$ we have $$(0,0,\dots,w_j,\dots,0)=\sum_{i=1}^{m}\lambda_i(u_{i1},u_{i2},\dots,u_{im}) = \left(\sum_{i=1}^{m}\lambda_{i}u_{i1},\sum_{i=1}^{m}\lambda_{i}u_{i2},\dots,\sum_{i=1}^{m}\lambda_{i}u_{im}\right)$$ implying $$w_j = \sum_{i=1}^{m}\lambda_{i}u_{ij}$$ and thus $w_j\in\operatorname{span}(u_{1j},u_{2j},\dots,u_{mj})$ and by extension $V_j\subseteq\operatorname{span}(u_{1j},u_{2j},\dots,u_{mj})$, and since $u_{1j},u_{2j},\dots,u_{mj}\in V_j$ it immediately follows that $\operatorname{span}(u_{1j},u_{2j},\dots,u_{mj})\subseteq V$.
$\blacksquare$