Without any further hypothesis on $U$, the answer is no. Take $V = \mathbb{C}^3$, $d=2$, $l=2$ and $U = S^2(W_1)$ where $W_1$ is a $\mathbb{C}^2$ inside $V$. Then for any other $W_2 \subset V$ of dimension $2$, we have:

$$S^2(W_1 \cap W_2) \subset U \cap S^2(W_2)$$

and obiously $\dim S^2(W_1 \cap W_2) \geq 1$. On the other hand, I think the answer should be yes if you assume $U$ to be generic (and perhaps some adequate numerical conditions). This should follow from a dimension count bounding the dimension of the subvariety of $\mathrm{Gr}\left( \binom{l+d-1}{l},S^d V \right)$ representing linear spaces which intersects $S^d(W)$ non-transversally for all $W \in \mathrm{Gr}(l,V)$.

Without any further hypothesis on $U$, the answer is no. Take $V = \mathbb{C}^3$, $d=2$, $l=2$ and $U = S^2(W_1)$ where $W_1$ is a $\mathbb{C}^2$ inside $V$. Then for any other $W_2 \subset V$ of dimension $2$, we have:

$$S^2(W_1 \cap W_2) \subset U \cap S^2(W_2)$$

and obiously $\dim S^2(W_1 \cap W_2) \geq 1$. On the other hand, I think the answer should be yes if you assume $U$ to be generic (and perhaps some adequate numerical conditions). This should follow from a dimension count bounding the dimension of the subvariety of $\mathrm{Gr}\left( \binom{l+d-1}{l},S^d V \right)$ representing linear spaces which intersect $S^d(W)$ non-transversally for all $W \in \mathrm{Gr}(l,V)$.