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This can fail. Consider the natural embedding $\operatorname{Sp}_4(\mathbb C) \to \operatorname{SL}_4(\mathbb C)$, where $\operatorname{Sp}_4(\mathbb C)$ is taken with respect to the symplectic form $(x_1, x_2, x_3, x_4) \mapsto x_1 x_4 - x_2 x_3$. (This is a map of complex Lie groups, but it carries maximal compact subgroups to maximal compact subgroups, so you can work in that setting if you prefer.) If we denote the simple (with respect to the upper-triangular Borel) roots of the diagonal torus in $\operatorname{SL}_4(\mathbb C)$ by $\alpha_1, \alpha_2, \alpha_3$, in the obvious fashion, then the restrictions of $\alpha_1$ and $\alpha_3$ to the diagonal torus in $\operatorname{Sp}_4(\mathbb C)$ are the short simple (with respect to the upper-triangular Borel) root $a$, and the restriction of $\alpha_2$ is the long simple root $b$. Then $a^\vee$ equals $\alpha_1^\vee + \alpha_3^\vee$, which is not a coroot of $\operatorname{SL}_4(\mathbb C)$.

(As discussed in comments, the original version of the question did not include the condition on centres, and so admitted any proper, maximal-rank subgroup as a counterexample. This example does satisfy the centre condition.)

This can fail. Consider the natural embedding $\operatorname{Sp}_4(\mathbb C) \to \operatorname{SL}_4(\mathbb C)$, where $\operatorname{Sp}_4$ is taken with respect to the symplectic form $(x, y) \mapsto x_1 y_4 + x_2 y_3 - x_3 x_2 - x_4 y_1$. (This is a map of complex Lie groups, but it carries maximal compact subgroups to maximal compact subgroups, so you can work in that setting if you prefer.) If we denote the simple (with respect to the upper-triangular Borel) roots of the diagonal torus in $\operatorname{SL}_4(\mathbb C)$ by $\alpha_1, \alpha_2, \alpha_3$, in the obvious fashion, then the restrictions of $\alpha_1$ and $\alpha_3$ to the diagonal torus in $\operatorname{Sp}_4(\mathbb C)$ are the short simple (with respect to the upper-triangular Borel) root $a$, and the restriction of $\alpha_2$ is the long simple root $b$. Then $a^\vee$ equals $\alpha_1^\vee + \alpha_3^\vee$, which is not a coroot of $\operatorname{SL}_4(\mathbb C)$.

(As discussed in comments, the original version of the question did not include the condition on centres, and so admitted any proper, maximal-rank subgroup as a counterexample. This example does satisfy the centre condition.)

This can fail. Consider the natural embedding $\operatorname{Sp}_4(\mathbb C) \to \operatorname{SL}_4(\mathbb C)$, where $\operatorname{Sp}_4(\mathbb C)$ is taken with respect to the symplectic form $(x_1, x_2, x_3, x_4) \mapsto x_1 x_4 - x_2 x_3$. (This is a map of complex Lie groups, but it carries maximal compact subgroups to maximal compact subgroups, so you can work in that setting if you prefer.) If we denote the simple (with respect to the upper-triangular Borel) roots of the diagonal torus in $\operatorname{SL}_4(\mathbb C)$ by $\alpha_1, \alpha_2, \alpha_3$, in the obvious fashion, then the restrictions of $\alpha_1$ and $\alpha_3$ to the diagonal torus in $\operatorname{Sp}_4(\mathbb C)$ are the short simple (with respect to the upper-triangular Borel) root $a$, and the restriction of $\alpha_2$ is the long simple root $b$. Then $a^\vee$ equals $\alpha_1^\vee + \alpha_3^\vee$, which is not a coroot of $\operatorname{SL}_4(\mathbb C)$.

In fact, less exotically, doesn't this fail for any proper, maximal-rank subgroup, for example, $\operatorname U_2 \times \operatorname U_1$ in $\operatorname U_3$, simply because $\mathfrak t$ equals $\mathfrak t'$, but $\Psi$ does not equal $\Psi'$?

This can fail. Consider the natural embedding $\operatorname{Sp}_4(\mathbb C) \to \operatorname{SL}_4(\mathbb C)$, where $\operatorname{Sp}_4(\mathbb C)$ is taken with respect to the symplectic form $(x_1, x_2, x_3, x_4) \mapsto x_1 x_4 - x_2 x_3$. (This is a map of complex Lie groups, but it carries maximal compact subgroups to maximal compact subgroups, so you can work in that setting if you prefer.) If we denote the simple (with respect to the upper-triangular Borel) roots of the diagonal torus in $\operatorname{SL}_4(\mathbb C)$ by $\alpha_1, \alpha_2, \alpha_3$, in the obvious fashion, then the restrictions of $\alpha_1$ and $\alpha_3$ to the diagonal torus in $\operatorname{Sp}_4(\mathbb C)$ are the short simple (with respect to the upper-triangular Borel) root $a$, and the restriction of $\alpha_2$ is the long simple root $b$. Then $a^\vee$ equals $\alpha_1^\vee + \alpha_3^\vee$, which is not a coroot of $\operatorname{SL}_4(\mathbb C)$.

(As discussed in comments, the original version of the question did not include the condition on centres, and so admitted any proper, maximal-rank subgroup as a counterexample. This example does satisfy the centre condition.)

This can fail. Consider the natural embedding $\operatorname{Sp}_4(\mathbb C) \to \operatorname{SL}_4(\mathbb C)$, where $\operatorname{Sp}_4(\mathbb C)$ is taken with respect to the symplectic form $(x_1, x_2, x_3, x_4) \mapsto x_1 x_4 - x_2 x_3$. (This is a map of complex Lie groups, but it carries maximal compact subgroups to maximal compact subgroups, so you can work in that setting if you prefer.) If we denote the simple (with respect to the upper-triangular Borel) roots of the diagonal torus in $\operatorname{SL}_4(\mathbb C)$ by $\alpha_1, \alpha_2, \alpha_3$, in the obvious fashion, then the restrictions of $\alpha_1$ and $\alpha_3$ to the diagonal torus in $\operatorname{Sp}_4(\mathbb C)$ are the short simple (with respect to the upper-triangular Borel) root $a$, and the restriction of $\alpha_2$ is the long simple root $b$. Then $a^\vee$ equals $\alpha_1^\vee + \alpha_3^\vee$, which is not a coroot of $\operatorname{SL}_4(\mathbb C)$.

I believe the statement is true if $T = T'$, or, nominally more generally, if $K$ is normalised by $T'$.

This can fail. Consider the natural embedding $\operatorname{Sp}_4(\mathbb C) \to \operatorname{SL}_4(\mathbb C)$, where $\operatorname{Sp}_4(\mathbb C)$ is taken with respect to the symplectic form $(x_1, x_2, x_3, x_4) \mapsto x_1 x_4 - x_2 x_3$. (This is a map of complex Lie groups, but it carries maximal compact subgroups to maximal compact subgroups, so you can work in that setting if you prefer.) If we denote the simple (with respect to the upper-triangular Borel) roots of the diagonal torus in $\operatorname{SL}_4(\mathbb C)$ by $\alpha_1, \alpha_2, \alpha_3$, in the obvious fashion, then the restrictions of $\alpha_1$ and $\alpha_3$ to the diagonal torus in $\operatorname{Sp}_4(\mathbb C)$ are the short simple (with respect to the upper-triangular Borel) root $a$, and the restriction of $\alpha_2$ is the long simple root $b$. Then $a^\vee$ equals $\alpha_1^\vee + \alpha_3^\vee$, which is not a coroot of $\operatorname{SL}_4(\mathbb C)$.

In fact, less exotically, doesn't this fail for any proper, maximal-rank subgroup, for example, $\operatorname U_2 \times \operatorname U_1$ in $\operatorname U_3$, simply because $\mathfrak t$ equals $\mathfrak t'$, but $\Psi$ does not equal $\Psi'$?