= Solution
First interpret the stated well-definedness as including uniqueness of the <global minimizer> for each positive parameter. Under that assumption, the <parameter continuity of variational regularization> follows from the <direct method in the calculus of variations> argument below.
Let $M=D(0)=\|f\|^2/2$. For large $n$, $\alpha_n\geq\beta:=\alpha/2$. Comparison with zero, using $J(0)=0$, gives
$$
\Phi_{\beta,f}(u_{\alpha_n})\leq\Phi_{\alpha_n,f}(u_{\alpha_n})\leq M,
\qquad J(u_{\alpha_n})\leq M/\beta.
$$
<Coercivity> of $\Phi_{\beta,f}$ therefore bounds the sequence. Take a $\tau$-<convergent subsequence> with limit $\bar u$. For any $v\in\operatorname{dom}J$, optimality gives
$$
\Phi_{\alpha,f}(u_{\alpha_n})
\leq\Phi_{\alpha,f}(v)
+(\alpha_n-\alpha)J(v)+(\alpha-\alpha_n)J(u_{\alpha_n}).
$$
The last two terms vanish. <Sequential lower semicontinuity> at the fixed parameter $\alpha$ implies $\Phi_{\alpha,f}(\bar u)\leq\Phi_{\alpha,f}(v)$. Thus every cluster point is a <global minimizer> at $\alpha$. Uniqueness makes it $u_\alpha$, and the subsequence property proves
$$
\boxed{u_{\alpha_n}\xrightarrow{\tau}u_\alpha.}
$$
Indeed, a subsequence staying outside a neighborhood of $u_\alpha$ would have a further convergent subsequence, whose limit must be $u_\alpha$, a contradiction.
\b[The listed existence hypotheses alone do not imply uniqueness or convergence of an arbitrary selection.] For example, on $\mathbb R^2$, set $K(x,y)=x$, $f=1$, and
$$
J(x,y)=x^2+\bigl(\max\{|y|-1,0\}\bigr)^2.
$$
This is nonnegative, <convex>, <continuous> and has $J(0)=0$; every $\Phi_{\alpha,f}$ has <coercivity>. Its minimizers are $(1/(1+2\alpha),y)$ for all $|y|\leq1$. Along $\alpha_n=\alpha+1/n$, selecting $y_n=(-1)^n$ prevents convergence. Without uniqueness, the valid conclusion is that every cluster point minimizes the limiting objective.
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