Need help with a proof - real analysis/epsilon delta

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What I need to prove is the following:

Given the continuous function $ℎ()$ on the closed interval $[,]$ and $y=\text{inf}\{x: h(x) \geq r\}$, and given the set is non-empty and $ \in [,)$, then $ℎ()=$.

My attempt at a proof: For $<$ in $[,]$, we have that $ℎ()<$. Thus, $ℎ(-\frac{1}{})<$. Letting $→∞$, we obtain $ℎ()≤$.

Now, there exists a decreasing sequence $_$ converging to $$ such that $ℎ()≥$. Taking the limit gives $ℎ()≥$. Thus $ℎ()=$, and we are done.

This proof is inspired by a proof from @KaviRamaMurthy, as a response to a previous thread of mine (Trying to understand a proof that continuity on a closed interval implies uniform continuity).

However, I'm not quite sure about this. An epsilon-delta proof using contradiction is apparently another approach, but I'm uncertain how to reach a contradiction.

I'm relatively new to analysis, so keep that in mind :).

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Your proof works if $y \in (a,b]$ otherwise if $y \in [a,b)$ you take the constant function $h(x)=r+1$ then $y=a$ and $h$ satisfy the hypothesis but $h(a)=h(y)=r+1$