$(a)$ Recall that reflexivity must hold FOR ALL subsets of a set, including $\varnothing$!
- $\varnothing \subset \mathbb R$.
- But $\varnothing \cap \varnothing = \varnothing$
Hence, since there exists a subset that doesn't satisfy reflexivity, the relation as a whole cannot be reflexive.
$(b)$ You are correct; the relation is symmetric: but it's symmetric because IF $x\cap y \neq \varnothing$, then $y\cap x = x \cap y \neq \varnothing.$
$(c)$ Recheck transitivity, too, to find a counterexample to the property: We want to show that if for any subsets $x, y, w \subseteq \mathbb R\,$ $\;x\,R\,y\;$ and $\;y\,R\,w,\;$ it follows that $\;x\,R\,w\;$ I think you got a little mixed up along the way as to *how the relation is defined. $\;x\,R\, y\;$ means $x\cap y\neq \varnothing$. (The intersection of subset $x$ and subset $y$ is non-empty.) And $\;y\, R \,w$ means $\;y\cap w \neq \varnothing.\;$ It does not necessarily follow that $x\,R\,w\;$, that is, there are counterexamples to $\;x \cap w \neq \varnothing.\;$
- E.g.: let $x, y, w$ be subsets of $\mathbb R$ defined by open intervals of reals: $x = (0, 2),\; y = (1, 3),\; w = (2, 4).\;$ Then $x \cap y = (1, 2) \neq \varnothing,\; y\cap w = (2, 3) \neq \varnothing,\;$ but $\;x\cap w = (0, 2) \cap (2, 4) = \varnothing$
$(d)$ In light of not being reflexive (1), refine your reason for not being irreflexive: There exist sets $x\subseteq \mathbb R$ such that $x \cap x\neq \varnothing$; and you need only provide an example of such a subset $x$ to show the relation is not irreflexive.