 Solution

It suffices first to treat unital algebras and a unital homomorphism. A surjective homomorphism between unital algebras automatically preserves the identity. If $B=0$, continuity is immediate. In the nonunital case, extend to the forced unitizations by $T^{+}(a+\lambda 1)=T(a)+\lambda 1$. This is still surjective and the unitization of a semisimple algebra is semisimple: its radical projects to zero in its scalar quotient, and its remaining intersection with the original algebra is its original radical. Continuity of the extension implies continuity of the original map.

An algebra homomorphism decreases the spectral radius, even without any continuity assumption, because it takes an inverse of $\lambda 1-a$ to an inverse of $\lambda 1-T(a)$. Suppose $a_n\to 0$ in $A$ and $T(a_n)\to b$ in $B$. Fix any $c\in B$ and choose $u,v\in A$ with $T(u)=b$, $T(v)=c$. PutConsider the polynomial $q_n(z)=T((1-z)w_n+zw)=(1-z)d_n+zd$. Its value at one is exactly $d$. On the outer circle $|z|=R$,On the inner circle $|z|=1/R$, spectral-radius decrease givesApplying part (i) in $B$ and then letting $n\to \infty$, with $R$ fixed, givesThis holds for every $R>1$, so $r_B(cb)=0$ for every $c\in B$.

This forces $b$ into the Jacobson radical. To see the criterion directly, if $b$ were outside a maximal left ideal $L$, then $L+Bb=B$, so $1=l+cb$ for some $l\in L$, $c\in B$. The equality $r_B(cb)=0$ makes $1-cb$ invertible, but this element equals $l\in L$, impossible in a proper left ideal. Hence $b$ lies in every maximal left ideal. Since $B$ is semisimple, $b=0$.

We have shown that the separating space of $T$ is zero. If $a_n\to a$ and $T(a_n)\to y$, apply the result to $a_n-a$ to obtain $y=T(a)$. Its graph is closed, and the closed graph theorem now provesThe crucial estimate uses part (i) with an exactly fixed central value; it does not assume the false general assertion that the spectral radius is continuous under norm limits.