As follow from comments to the question, you thought that, in the spirit of 't Hooft's anomaly matching conditions, anomalies, which break the symmetry explicitly, leads to existence of Nambu-Goldstone bosons. This is misunderstanding of conditions.

Really, 't Hooft anomaly matching says that if we initially have theory with chiral fermions and gauge bosons, and feemion sector has the global symmetry group $G$, which is gauge anomaly free (no $G_{\text{gauge}}^2G$ anomalies), but which is anomalous itself (there exist $G^3$ anomalies), then after the confinement there must exist massless fermion bound states which reproduce anomalous structure of symmetry group in initial theory; otherwise there must exist spontaneous symmetry breaking. But is there symmetry breaking (i.e., current nonconservation) by anomaly?

In fact, no. Presence of $G^3$ anomalies means only that there are symmetric tensor with group indices structure in the triangle diagram with three $G$ currents, namely $$ d_{abc} \equiv \text{Tr}[[t_{a},t_{b}]_{+}t_{c}] $$ This means that if we decide to gauge the $G$ group, we will face with current nonconservation. Otherwise there is no current nonconservation and hence no symetry breaking. For example, the pure QCD global symmetry group $SU_{L}(3)\times SU_{R}(3)\times U_{B}(1)$ is exact in the chiral limit, although it is anonalous itself.

First note that chiral symmetry breaking appears at the level of states of theory, where exist set of nontrivial vacuums which are connected with each other by symmmetry transformations, and hence which are not invariant under the gauge transformation. The theory itself on the level of path integral is gauge invariant. If the theory isn't invariant, then we may talk about SSB of approximately exact symmetry, if breaking effects are small compared to themscale of CSB, otherwise there is no even approximate symmetry.

As follow from comments to the question, you thought that, in the spirit of 't Hooft's anomaly matching conditions, anomalies, which break the symmetry explicitly, leads to existence of Nambu-Goldstone bosons. This is misunderstanding of conditions.

Really, 't Hooft anomaly matching says that if we initially have theory with chiral fermions and gauge bosons, and feemion sector has the global symmetry group $G$, which is gauge anomaly free (no $G_{\text{gauge}}^2G$ anomalies), but which is anomalous itself (there exist $G^3$ anomalies), then after the confinement there must exist massless fermion bound states which reproduce anomalous structure of symmetry group in initial theory; otherwise there must exist spontaneous symmetry breaking. But is there symmetry breaking (i.e., current nonconservation) by anomaly?

In fact, no. Presence of $G^3$ anomalies means only that there are symmetric tensor with group indices structure in the triangle diagram with three $G$ currents, namely $$ d_{abc} \equiv \text{Tr}[[t_{a},t_{b}]_{+}t_{c}] $$ This means that if we decide to gauge the $G$ group, we will face with current nonconservation. Otherwise there is no current nonconservation and hence no symetry breaking. For example, the pure QCD global symmetry group $G \sim SU_{L}(3)\times SU_{R}(3)\times U_{B}(1)$ is exact in the chiral limit, although it is anonalous itself.

As follow from comments to the question, you thought that, in the spirit of 't Hooft, anomaly matching conditions anomalies, which break the symmetry explicitly, leads to existence of Nambu-Goldstone bosons. This is misunderstanding of conditions.

Really, 't Hooft anomaly matching says that if we initially have theory with chiral fermions and gauge bosons, and feemion sector has the global symmetry group $G$, which is gauge anomaly free (no $G_{\text{gauge}}^2G$ anomalies), but which is anomalous itself (there exist $G^3$ anomalies), then after the confinement there must exist massless fermion bound states which reproduce anomalous structure of symmetry group in initial theory; otherwise there must exist spontaneous symmetry breaking. But is there symmetry breaking (i.e., current nonconservation) by anomaly?

In fact, no. Presence of $G^3$ anomalies means only that there are symmetric tensor with group indices structure in the triangle diagram with three $G$ currents, namely $$ d_{abc} \equiv \text{Tr}[[t_{a},t_{b}]_{+}t_{c}] $$ This means that if we decide to gauge the $G$ group, we will face with current nonconservation. Otherwise there is no current nonconservation and hence no symetry breaking. For example, the pure QCD global symmetry group $SU_{L}(3)\times SU_{R}(3)\times U_{B}(1)$ is exact in the chiral limit, although it is anonalous itself.

As follow from comments to the question, you thought that, in the spirit of 't Hooft's anomaly matching conditions, anomalies, which break the symmetry explicitly, leads to existence of Nambu-Goldstone bosons. This is misunderstanding of conditions.

Really, 't Hooft anomaly matching says that if we initially have theory with chiral fermions and gauge bosons, and feemion sector has the global symmetry group $G$, which is gauge anomaly free (no $G_{\text{gauge}}^2G$ anomalies), but which is anomalous itself (there exist $G^3$ anomalies), then after the confinement there must exist massless fermion bound states which reproduce anomalous structure of symmetry group in initial theory; otherwise there must exist spontaneous symmetry breaking. But is there symmetry breaking (i.e., current nonconservation) by anomaly?

In fact, no. Presence of $G^3$ anomalies means only that there are symmetric tensor with group indices structure in the triangle diagram with three $G$ currents, namely $$ d_{abc} \equiv \text{Tr}[[t_{a},t_{b}]_{+}t_{c}] $$ This means that if we decide to gauge the $G$ group, we will face with current nonconservation. Otherwise there is no current nonconservation and hence no symetry breaking. For example, the pure QCD global symmetry group $SU_{L}(3)\times SU_{R}(3)\times U_{B}(1)$ is exact in the chiral limit, although it is anonalous itself.