Brezis' exercise 8.24.2: do we need the assumption that $k$ is sufficiently large?

Let $$I$$ be the open interval $$(0, 1)$$. I'm trying to solve a problem in Brezis' Functional Analysis, i.e.,

Exercise 8.24

1. Prove that for every $$\varepsilon>0$$ there exists a constant $$C_{\varepsilon}$$ such that $$|u(1)|^2 \leq \varepsilon\left\|u^{\prime}\right\|_{L^2}^2+C_{\varepsilon}\|u\|_{L^2}^2 \quad \forall u \in H^1(I) .$$
2. Prove that if the constant $$k>0$$ is sufficiently large, then for every $$f \in L^2(I)$$ there exists a unique $$u \in H^2(I)$$ satisfying $$(1) \quad \begin{cases} -u^{\prime \prime}+k u=f \quad \text{on} \quad I, \\ u^{\prime}(0)=0 \quad \text{and} \quad u^{\prime}(1)=u(1). \end{cases}$$ What is the weak formulation of problem (1)? What is the associated minimization problem?

In below attempt of (2.), I don't need that $$k$$ is sufficiently large. I only need $$k \ge 0$$. Could you verify if my attempt contains some subtle mistakes?

Let $$K := \{v \in H^1(I) : v'(0)=0, v'(1)=v(1)\}$$. Then $$K$$ is a closed subspace of $$H^1(I)$$. If $$u$$ is a classical solution to $$(1)$$, then $$\int_I [-u''v + kuv] = \int_I fv, \quad \forall v \in K,$$ which (by integration by parts) is equivalent to $$(2) \quad u(1) v(1) + \int_I [u'v' + k uv] = \int_I fv, \quad \forall v \in K.$$

Then $$(2)$$ is the weak forlulation of $$(1)$$. We define a symmetric bilinear form $$a$$ on $$K$$ by $$a(u, v) := u(1) v(1) + \int_I [u'v' + k uv].$$

It follows from (1.) that $$a$$ is continuous. Because $$k$$ is non-negative, $$a$$ is coercive. We define $$\varphi \in (H^1(I))^*$$ by $$\varphi (v) = \int_I fv, \quad \forall v \in K.$$

By Lax-Milgram theorem, $$(2)$$ has a unique solution $$u \in K$$. The associated minimization is $$u= \operatorname{argmin}_{v \in K} \left \{ \frac{1}{2} a(v, v) - \varphi (v)\right \}.$$

Notice that $$(2)$$ implies $$\int_I u'v' = -\int_I (ku-f)v, \quad \forall v \in C^\infty_c (I),$$ which implies $$u \in H^2(I)$$ with $$u''=ku-f$$.

Your definition of $$K$$ is not well-defined because it involves $$v'(0)$$ and $$v'(1)$$ while we only know $$v \in H^1 (I)$$. The equation $$(2)$$ is also wrong because the wrong sign of $$u(1) v(1)$$, and this is where we need $$k$$ to be large enough. Below is a fix.
If $$u$$ is a classical solution to $$(1)$$, then $$\int_I [-u''v + kuv] = \int_I fv, \quad \forall v \in H^1 (I),$$ which (by integration by parts) implies $$(2) \quad -u(1) v(1) + \int_I [u'v' + k uv] = \int_I fv, \quad \forall v \in H^1 (I).$$
Then $$(2)$$ is the weak forlulation of $$(1)$$. We define a symmetric bilinear form $$a$$ on $$H^1(I)$$ by $$a(u, v) := -u(1) v(1) + \int_I [u'v' + k uv].$$
It follows from (1.) that $$a$$ is continuous and that if $$k>0$$ is sufficiently large then $$a$$ is coercive. We define $$\varphi \in (H^1(I))^*$$ by $$\varphi (v) = \int_I fv, \quad \forall v \in K.$$
By Lax-Milgram theorem, $$(2)$$ has a unique solution $$u \in H^1(I)$$. The associated minimization is $$u= \operatorname{argmin}_{v \in K} \left \{ \frac{1}{2} a(v, v) - \varphi (v)\right \}.$$
Notice that $$(2)$$ implies $$\int_I u'v' = -\int_I (ku-f)v, \quad \forall v \in C^\infty_c (I),$$ which implies $$u \in H^2(I)$$ with $$u''=ku-f$$. By integration by parts, $$(2)$$ implies $$(3) \quad (u'(1)-u(1))v(1) -u'(0) v(0) =0, \quad \forall v \in H^1 (I),$$ which implies $$u'(1)=u(1)$$ and $$u'(0)=0$$.