Suppose we are given a pushout diagram of simplicial sets $\sigma :$
\[ \xymatrix@R =50pt@C=50pt{ S \ar [r] \ar [d] & T \ar [d] \\ S' \ar [r] & T', } \]
where the horizontal maps are monomorphisms. Suppose that $S$, $T$, and $S'$ are good; we wish to show that $T'_{\bullet }$ is good. Note that the horizonal maps induce monomorphisms of directed graphs
\[ G(S) \hookrightarrow G(T ) \quad \quad G(S' ) \hookrightarrow G(T'). \]
Define subgraphs $G_0(S) \subseteq G(S )$ and $G_0( T) \subseteq G(T )$ by the formulae
\[ \operatorname{Vert}( G_0(S ) ) = \operatorname{Vert}( G(S) ) = S_0 \quad \quad \operatorname{Vert}( G_0(T ) ) = \operatorname{Vert}( G(T) ) = T_0 \]
\[ \operatorname{Edge}( G_0( S) ) = \emptyset \quad \quad \operatorname{Edge}( G_0( T ) ) = \operatorname{Edge}( G(T) ) \setminus \operatorname{Edge}( G( S) ). \]
We then have a commutative diagram of categories
\[ \xymatrix@R =50pt@C=50pt{ \operatorname{Path}[ G_0(S)] \ar [r] \ar [d] & \operatorname{Path}[ G_0(T) ] \ar [d] \\ \operatorname{Path}[ G( S' ) ] \ar [r] \ar [d]^{ F_{S'} } & \operatorname{Path}[ G( T' )] \ar [d]^{ F_{T'} } \\ \operatorname{Path}[ S' ]_{m} \ar [r] & \operatorname{Path}[ T' ]_{m}. } \]
We wish to show that the functor $F_{T'}$ is an isomorphism of categories, and the map $F_{S' }$ is an isomorphism by assumption. It will therefore suffice to show that the lower square in this diagram is a pushout. Note that the upper square is a pushout (since it is obtained from a pushout diagram in the category of directed graphs by passing to path categories). We are therefore reduced to showing that the outer rectangle is a pushout. We can rewrite this as the outer rectangle in another commutative diagram of categories
\[ \xymatrix@R =50pt@C=50pt{ \operatorname{Path}[ G_0(S)] \ar [r] \ar [d] & \operatorname{Path}[ G_0(T) ] \ar [d] \\ \operatorname{Path}[ G( S ) ] \ar [r] \ar [d]^{ F_{S} } & \operatorname{Path}[ G(T )] \ar [d]^{ F_{T} } \\ \operatorname{Path}[ S ]_{m} \ar [r] \ar [d] & \operatorname{Path}[ T ]_{m} \ar [d] \\ \operatorname{Path}[ S' ]_{m} \ar [r] & \operatorname{Path}[ T' ]_{m}. } \]
We now conclude by observing that the upper square in this diagram is a pushout (because it is obtained from a pushout diagram of directed graphs by passing to path categories), the middle square is a pushout (since $F_{S}$ and $F_{T}$ are isomorphisms), and the lower square is a pushout (since the construction $X_{\bullet } \mapsto \operatorname{Path}[ X ]_{m}$ preserves colimits).