Workaround:
First we solve PDE only with first two ic's.
$Version
(*12.1.0 for Microsoft Windows (64-bit) (March 14, 2020)*)
Heat = D[T[y, t], t] - α*D[T[y, t], {y, 2}] == 0;
ic = {T[y, 0] == 0, T[0, t] == 1};
sol = DSolve[{Heat, ic}, T[y, t], {y, t}, Assumptions -> {α > 0, t > 0, y > 0}]
(*{{T[y, t] -> (
y Inactive[Integrate][
E^(-(y^2/(4 t α - 4 α K[2])))/(t - K[2])^(
3/2), {K[2], 0, t}, Assumptions -> True])/(
2 Sqrt[π] Sqrt[α])}}*)
sol // Activate (* Mathematica Can't find Integral ? *)
(*{{T[y, t] -> (
y Inactive[Integrate][
E^(-(y^2/(4 t α - 4 α K[2])))/(t - K[2])^(
3/2), {K[2], 0, t}, Assumptions -> True])/(
2 Sqrt[π] Sqrt[α])}}*)
Workaround to solve integral:
FullSimplify[
InverseMellinTransform[
Integrate[
MellinTransform[(
y E^(-((A*y^2)/(4 t α - 4 α K[2])))/(t - K[2])^(
3/2))/(2 Sqrt[π] Sqrt[α]), A, s], {K[2], 0, t},
Assumptions -> {t > 0, α > 0, y > 0}], s, A] /. A -> 1,
Assumptions -> {t > 0, y > 0, α > 0}] // Expand
(*Erfc[y/(2 Sqrt[t α])]*)
We check if this solution fulfills Boundary Condition
$T(\infty ,t)=0$
Limit[Erfc[y/(2 Sqrt[t α])], y -> Infinity,
Assumptions -> {α > 0, y > 0, t > 0}]
(* 0 *) (*Check OK*)
We check if this solution fulfills Initial Condition
$T(0 ,t)=1$
Limit[Erfc[y/(2 Sqrt[t α])], y -> 0, Assumptions -> {t > 0, α > 0}]
(*1 *)
We check if this solution fulfills Initial Condition
$T(y ,0)=0$
Limit[Erfc[y/(2 Sqrt[t α])], t -> 0, Assumptions -> {y > 0, α > 0}, Direction -> -1]
(* 0 *)
Check if solution is true:
Heat /. {T -> Function[{y, t}, Erfc[y/(2 Sqrt[t α])]]}
(*True*)
Then solution is:
$$T(y,t)=\text{erfc}\left(\frac{y}{2 \sqrt{t \alpha }}\right)$$