Relative compactness (closure is compact) of a set in a Skorokhod space is characterized by for instance Theorem 12.3 in Billingsley's ''Convergence of probability measures''. But it is not clear when such a relative compact set is also closed in the Skorokhod topology or what its closure looks like.
A simpler question to start would be maybe: Let us look at the space $D$ of cadlag functions from $[0,1]$ to $\mathbb{R}$ equipped with the Skorokhod topology. Is the set {$\{f\in D\colon \sup_x |f(x)| \leq 2\}$} closed in the Skorokhod topology? If not what is its closure?
The first question is too general. Closures may be different.
The answer for the second one: $A = \{f\in D\colon \sup_x |f(x)| \leq 2\}$ is closed. Let's prove it. Suppose that $f \notin A$. Hence $f(x_0) > 2$ at $x \in [0,1]$. We will suppose what $x_0 < 1$ - otherwise everything is easier. Put $\varepsilon = \frac{f(x_0)-2}{3} > 0$. Thus $f(x_0) > 2 + 3 \varepsilon$. As $f(x)$ is right-continious, then there exists $0 < \delta < \varepsilon$ such that $|f(x) - f(x_0)| < \varepsilon $ for $x \in [x_0, x_0 + 2\delta]$. We got that $f(x) > 2 + 2\varepsilon$ for $x \in [x_0, x_0 + 2\delta]$.
Now consider $g$ such that $|| g - f|| < \frac{\delta}{2}$. Then by definition there exists $\lambda: [0,1]\to[0,1]$ such that $\sup_t |\lambda t - t | \le \delta$ and $\sup_t |g(t) - f(\lambda t)| \le \delta$. Hence $\lambda (x_0 + \delta) \in [x_0, x_0 + 2\delta]$. Thus $$f(\lambda (x_0 + \delta)) \ge 2 + 2\varepsilon \quad (*).$$ But $|g(x_0 + \delta) - f(\lambda (x_0 + \delta)) | \le \delta < \varepsilon$ and it follows from $(*)$ that $g(x_0 + \delta) > 2 + \varepsilon$. Thus $g \notin A$. We showed that the complement of $A$ is open. Q.e.d.
It's a formalization of intuitive proof: we see that if $f(x) > 2$ at some point then if we make a little modification of $f$ in $x$-axis and $y$-axis then the modified function $g$ will still be bigger than $2$ at some point.