Choose a smooth surjective morphism $f:U \to \mathcal{X}$ with source a scheme, necessarily reduced since $\mathcal{X}$ is, and write $U' = \mathcal{T}\times_{\mathcal{X}} U$.  The base-changed morphism $U' \to U$ is a monomorphism of algebraic spaces, locally of finite type, and thus representable Morphisms of Spaces, Lemma \ref{spaces-morphisms-lemma-locally-quasi-finite-separated-representable} and \ref{spaces-morphisms-lemma-monomorphism-loc-finite-type-loc-quasi-finite}; since $U$ is a scheme, so is $U'$. The projection $f': U' \to \mathcal{T}$ is again a smooth surjection. Let $u' \in U'$, with image $u \in U$. Lemma \ref{lemma-base-change-invariance-of-relative-dimension} shows that $\dim_{u'}(U'_{f(u')}) = \dim_u(U_{f(u)}),$ while $\dim_{f'(u')}(\mathcal{T}) =d \geq \dim_{f(u)}(\mathcal{X})$ by Lemma \ref{lemma-irreducible-implies-equidimensional} and our assumptions on $\mathcal{T}$ and $\mathcal{X}$. Thus we see that \begin{equation} \label{equation-dim-inequality} \dim_{u'} (U') = \dim_{u'} (U'_{f(u')}) + \dim_{f'(u')}(\mathcal{T}) \\ \geq \dim_u (U_{f(u)}) + \dim_{f(u)}(\mathcal{X}) = \dim_u (U). \end{equation} Since $U' \to U$ is a monomorphism, locally of finite type, it is in particular unramified, and so by the \'etale local structure of unramified morphisms \'Etale Morphisms, Lemma \ref{etale-lemma-finite-unramified-etale-local}, we may find a commutative diagram $$ \xymatrix{ V' \ar[r]\ar[d] & V \ar[d] \\ U' \ar[r] & U } $$ in which the scheme $V'$ is non-empty, the vertical arrows are \'etale, and the upper horizontal arrow is a closed immersion. Replacing $V$ by a quasi-compact open subset whose image has non-empty intersection with the image of $U'$, and replacing $V'$ by the preimage of $V$, we may further assume that $V$ (and thus $V'$) is quasi-compact. Since $V$ is also locally Noetherian, it is thus Noetherian, and so is the union of finitely many irreducible components. \medskip\noindent Since \'etale morphisms preserve pointwise dimension Descent, Lemma \ref{descent-lemma-dimension-at-point-local} we deduce from (\ref{equation-dim-inequality}) that for any point $v' \in V'$, with image $v \in V$, we have $\dim_{v'}( V') \geq \dim_v(V)$. In particular, the image of $V'$ can't be contained in the intersection of two distinct irreducible components of $V$, and so we may find at least one irreducible open subset of $V$ which has non-empty intersection with $V'$; replacing $V$ by this subset, we may assume that $V$ is integral (being both reduced and irreducible).  From the preceding inequality on dimensions, we conclude that the closed immersion $V' \hookrightarrow V$ is in fact an isomorphism. If we let $W$ denote the image of $V'$ in $U'$, then $W$ is a non-empty open subset of $U'$ (as \'etale morphisms are open), and the induced monomorphism $W \to U$ is \'etale (since it is so \'etale locally on the source, i.e.\ after pulling back to $V'$), and hence is an open immersion (being an \'etale monomorphism). Thus, if we let $\mathcal{V}$ denote the image of $W$ in $\mathcal{T}$, then $\mathcal{V}$ is a dense (equivalently, non-empty) open substack of $\mathcal{T}$, whose image is dense in an irreducible component of $\mathcal{X}$. Finally, we note that the morphism is $\mathcal{V} \to \mathcal{X}$ is smooth (since its composite with the smooth morphism $W\to \mathcal{V}$ is smooth), and also a monomorphism, and thus is an open immersion.