I'm trying to understand a step in the forward direction of the proof of the theorem $\lim_{n \to \infty} a_n = b \iff \limsup_{n \to \infty} a_n = \liminf_{n \to \infty} a_n = b$.
First, to clarify, let $\{a_n\}_{n = 1}^\infty \subseteq \mathbb{R}$. We define $M_n := \sup\{a_k : k \geq n\}$ and $m_n := \inf\{a_k: k \geq n\}$ yielding two new sequences $\{M_n\}$ and $\{m_n\}$ which may contain the symbols $\infty$ or $-\infty$. We then define $\limsup_{n \to \infty}a_n := \lim_{n \to \infty}M_n$ and $\liminf_{n \to \infty}a_n := \lim_{n \to \infty} m_n$ which may again be $\infty$ or $-\infty$.
Proof: ($\Longrightarrow$) [We limit this proof to limits converging to $b \in \mathbb{R}$ for now.] Suppose that $\lim_{n \to \infty}a_n = b$. For any $\varepsilon > 0$ there exists $N \in \mathbb{Z}^+$ such that $|a_n - b| < \varepsilon$ for all $n \geq N$. Those, we have $b - \varepsilon < a_n < b + \varepsilon$. But then, $\mathbf{b - \varepsilon < M_n \leq b + \varepsilon}$ and $\mathbf{b - \varepsilon \leq m_n < b + \varepsilon}$ for all $\mathbf{n \geq N}$. Since this holds for every $\varepsilon > 0$, we have $\limsup_{n \to \infty} a_n = \liminf_{n \to \infty} a_n = b$.
I am not sure how the bold statement follows immediately. Clearly, $b - \varepsilon < M_n$ and $m_n < b + \varepsilon$, but I don't see why $M_n \leq b + \varepsilon$ and $b - \varepsilon \leq m_n$. I am sure it's something simple that I'm not seeing (properties of supremum and infimum probably), but any help would be appreciated.