Given an infinite poset of a certain cardinality, does it contains always a chain or antichain of the same cardinality? It is known that given an infinite poset, it always contains an infinite chain or antichain;
moreover, there is a constructive proof that we can find a continuous chain in $P(\mathbb{N})$;
so, in general, I'm asking if given a poset of a certain cardinality, we could always find a chain or antichain of the same cardinality.
 A: A Suslin tree is a poset (in fact, a tree) that has cardinality $\omega_1$ but every chain and every antichain is countable. The existence of a Suslin tree is neither provable nor disprovable in ZFC. Therefore, your question does not have an affirmitive answer provable in ZFC. 
I do not know whether it has a negative answer (using something other than Suslin trees) provable in ZFC. Aubrey da Cunha has given a proof that the answer to the question is "no". That answer should be accepted over this one. 
A: A natural negative answer is to take the poset $\bigcup_{n\in\omega} \{n\}\times\omega_n$, with two points comparable iff their first coordinates coincide. Any antichain here is countable, and any chain has size strictly below $\aleph_\omega$, which is the size of the whole set. Note that this does not use any form of choice. 
A: In fact, there is a negative answer provable in ZFC.  Since we have choice, well-order the interval $[0,1]$.  Then let $x\sqsubset y$ if and only if the well-order agrees with the standard order of the reals and $x<y$.  Then $[0,1]$ with the ordering $\sqsubset$ is an uncountable poset with neither an uncountable chain nor an uncountable anti-chain.
Consider any uncountable $S\subseteq[0,1]$.  Let $z$ be the infimum of the set $$\{x\in [0,1]|\text{ there are uncountably many }y<x\text{ with }y\in S\}$$  $S$ cannot be a chain since otherwise, a countable increasing sequence of elements of $S$ converging to $z$ would be cofinal in $\omega_1$.  
Similarly, we can take $w$ to be the supremum of the set $$\{x\in [0,1]|\text{ there are uncountably many }y>x\text{ with }y\in S\}$$  Then if $S$ was an anti-chain, then a countable decreasing sequence (according to the standard ordering of the reals) in $S$ would again be cofinal in $\omega_1$.
A: This is true exactly for $\omega$ and for weakly compact cardinals. It is proved in 

G.D. Badenhorst and T. Sturm, An order- and graph-theoretical characterisation of weakly compact cardinals, in Cycles and Rays (NATO Adv. Sci. Inst. Ser. C Math. Phys. Sci., 301), 1990, pp.19-20, MR1096981.

A: This is the same answer as @aubrey's, but with an alternate proof.
Let $\preceq$ well-order the real numbers $\mathbb R$, and let $\leq$ be the usual ordering of $\mathbb R$. Then $\sqsubseteq$ defined by $x\sqsubseteq y$ if and only if $x\preceq y$ and $x\leq y$ is a partial order of $\mathbb R$.
We claim that every chain and antichain with respect to $\sqsubseteq$ is countable. To see this, let $C=\{c_\alpha:\alpha<|C|\}$ such that $\alpha<\beta$ if and only if $c_\alpha\prec c_\beta$.
We use the following lemma: given any increasing/decreasing sequence of reals, there is an increasing/decreasing sequence of rational numbers of the same cardinality (choose a rational between each term and its successor). Therefore every increasing/decreasing sequence of reals is countable.
If $C$ is a chain, then $\alpha<\beta$ implies $c_\alpha < c_\beta$, since $c_\alpha$ and $c_\beta$ are comparable by $\sqsubseteq$. Since there does not exist an uncountable increasing sequence of reals, $C$ is countable.
If $C$ is an antichain, then $\alpha<\beta$ implies $c_\alpha \not< c_\beta$ (equivalently, $c_\alpha >c_\beta$), since $c_\alpha$ and $c_\beta$ are incomparable by $\sqsubseteq$. Since there does not exist an uncountable decreasing sequence of reals, $C$ is countable.
