 Solution

True. Choose a basis $u_1,\dots ,u_r$ of $U$ and extend it to a basis $u_1,\dots ,u_r,w_1,\dots ,w_s$ of the finite-dimensional vector space $V$. Put $W=\mathrm{span}(w_1,\dots ,w_s)$. Every basis vector belongs to $U+W$, so $U+W=V$. Linear independence of the complete basis makes a vector belonging to both spans necessarily zero. ThereforeThis constructs a direct-sum complement; it need not be unique.