Poset dimension and width (Dilworth's theorem)

For your modified question, namely, does there exist a poset $P$ such that $\dim(P) > \sup \{|A|: A\subset P \text{ is an antichain}\}$, the answer is positive. Due to Laver (An Order Type Decomposition Theorem, Annals of Mathematics Second Series, Vol. 98, No. 1 (Jul., 1973), pp. 96-119) and in fact he showed that

for any uncountable $\kappa$, there exist a poset $P$ whose width is $\aleph_0$ and dimension is $\kappa$.

The poset is the class of scattered linear orders of size $<\kappa$ ordered by embeddability (modulo bi-embeddability to make it a poset).

This in some sense is a strengthening of Perles' result.

Let $P=P_1\cup P_2\cup P_3\cup\cdots$ where $P_1\lt P_2\lt P_3\lt\cdots$, each $P_n$ is finite, and $\dim(P_n)=n$. Then $\dim(P)=\operatorname{width}(P)=\aleph_0$.

More generally, if $\kappa$ is a strong limit cardinal, there is a poset $P$ with $\dim(P)=\operatorname{width}(P)=|P|=\kappa$.

Maybe I should have posted this as a comment rather than an answer. All these examples show is that you're using the wrong definition of "width". Try defining the width naturally, as the supremum of the cardinalities of antichains; and then ask whether there are any counterexamples to $\dim(P)\le\operatorname{width}(P)$. I have no idea.