The following proofs are influenced by Halmos's Measure Theory, Rudin's Real and Complex Analysis, Hewitt & Ross's Abstract Harmonic Analysis I and Folland's Real Analysis.
Definition
For an open subset $U$ of $X$, we write $\mu(U) = sup \{\lambda(K): K \subset U, K \in \Gamma \}$.
For any subset $E$ of $X$, we write $\mu(E) = inf \{\mu(U): E \subset U, U open \}$.
Lemma 1
$\mu(K) = \lambda(K)$ whenever $K \in \Gamma$.
Proof:
Choose $\epsilon \gt 0$. There exists $L\in \Gamma$ such that $K \subset L^{\circ}$ and $\lambda(L) - \lambda(K) \lt \epsilon$.
Since $\lambda(K) \le \mu(K) \le \mu(L^{\circ}) \le \lambda(L)$ and $\epsilon$ can be arbitrarily small, $\mu(K) = \lambda(K)$.
Lemma 2
Let $C \subset U \cup V$ where $C$ is compact and $U$ and $V$ are open.
Then there exist compact sets $K$ and $L$ such that $C = K \cup L$ and $K\subset U$, $L\subset V$.
Proof:
Since $C - U$ and $C - V$ are disjoint compact sets, There exist disjoint open sets $U_1$ and $V_1$ such that $C - U \subset U_1$ and $C - V \subset V_1$. Put $K = C - U_1$ and $L = C - V_1$.
Clearly $K \subset U$ and $L \subset V$. Since $K \cup L = (C - U_1) \cup (C - V_1) = C - (U_1\cap V_1) = C$, we are done.
Lemma 3
$\mu(U_1\cup U_2)\le \mu(U_1) + \mu(U_2)$ whenever $U_1$ and $U_2$ are open sets.
Proof:
Suppose $K \subset U_1 \cup U_2$, where $K\in \Gamma$.
By Lemma 2, there exist compact sets $C_1$ and $C_2$ such that $K = C_1\cup C_2$, $C_1\subset U_1$, $C_2\subset U_2$. By the assumption, there exist $K_1, K_2\in \Gamma$ such that $C_1\subset K_1\subset U_1$, $C_2\subset K_2\subset U_2$. Since $K\subset K_1\cup K_2$, $\lambda(K) \le \lambda(K_1\cup K_2)\le \lambda(K_1) + \lambda(K_2)$. Hence $\lambda(K) \le \mu(U_1) + \mu(U_2)$.
Hence $\mu(U_1\cup U_2) \le \mu(U_1) + \mu(U_2)$.
Lemma 4
$\mu(U_1\cup U_2) = \mu(U_1) + \mu(U_2)$ whenever $U_1$ and $U_2$ are disjoint open sets.
Proof:
Suppose $K_1\subset U_1$ and $K_2\subset U_2$, where $K_1, K_2\in \Gamma$.
Since $K_1$ and $K_2$ are disjoint and $K_1\cup K_2\in \Gamma$, $\lambda(K_1) + \lambda(K_2) = \lambda(K_1\cup K_2) \le \mu(U_1\cup U_2)$. Hence $\mu(U_1) + \mu(U_2) \le \mu(U_1 \cup U_2)$.
On the other hand, by Lemma 3, $\mu(U_1\cup U_2)\le \mu(U_1) + \mu(U_2)$ and we are done.
Lemma 5
Let $E_1, E_2, \cdots$ be a sequence of subsets of $X$.
Then $\mu(\cup E_n) \le \Sigma \mu(E_n)$.
Proof:
We may assume $\mu(E_n) \lt \infty$ for all $n$.
Choose $\epsilon \gt 0$. There exist open sets $U_n \supset E_n$ such that $\mu(U_n) \lt \mu(E_n) + \epsilon/2^n, n = 1,2,3,\cdots$.
Put $U = \cup U_n$. Suppose $K\subset U, K\in \Gamma$.
Since $K$ is compact, $K\subset U_1\cup \cdots \cup U_m$ for some $m$.
By Lemma 3, $\lambda(K) \le \mu(U_1\cup \cdots \cup U_m) \le \mu(U_1) + \cdots \mu(U_m) \le \Sigma \mu(E_n) + \epsilon$.
Hence $\mu(U) \le \Sigma \mu(E_n) + \epsilon$.
Since $\cup E_n \subset U$, $\mu(\cup E_n) \le \mu(U) \le \Sigma \mu(E_n) + \epsilon$. Hence $\mu(\cup E_n) \le \Sigma \mu(E_n)$.
Definition
A subset $E$ of $X$ is said to be $\mu$-measurable if it satisfies the following condition.
For every subset $S$ of $X$, $\mu(S) \ge \mu(S\cap E) + \mu(S-E)$.
We denote the family of $\mu$-measurable sets by $\mathcal M$.
Lemma 5 shows that $\mu$ is a Caratheodory outer measure defined on the power set of $X$.
It is well-known that $\mathcal M$ is a $\sigma$-algebra and $\mu$ is countably additive on it.
See for example Halmos's Measure Theory for the proofs.
Lemma 6
A subset $E$ of $X$ is $\mu$-measurable if
$\mu(U) \ge \mu(U\cap E) + \mu(U - E)$
for all open sets $U$ such that $\mu(U) \lt \infty$.
Proof:
Let $S$ be any subset of $X$.
We need to prove that $\mu(S) \ge \mu(S\cap E) + \mu(S-E)$.
If $\mu(S) = \infty$, this is trivially true. So we may assume $\mu(S) \lt \infty$.
Let $U$ be an open set such that $U \supset S$ and $\mu(U) \lt \infty$.
Then we have $\mu(U) \ge \mu(U\cap E) + \mu(U - E) \ge \mu(S\cap E) + \mu(S-E)$.
Hence $\mu(S) \ge \mu(S\cap E) + \mu(S-E)$.
Lemma 7
Every open subsets of $X$ is $\mu$-measurable.
Proof:
Let $V$ be an open subset of $X$.
By Lemma 6, it suffices to prove that $\mu(U) \ge \mu(U\cap V) + \mu(U - V)$
for every open set $U$ such that $\mu(Uļ¼\lt\infty$.
Choose $\epsilon\gt 0$.
There exists $K \in \Gamma$ such that $K \subset U \cap V$ and $\lambda(K) \gt \mu(U \cap V) - \epsilon$. Since $U - K$ is open, there exists $L \in \Gamma$ such that $L \subset U - K$ and $\lambda(L) \gt \mu(U - K) - \epsilon$. Since $K \cup L \subset U$ and $K\cap L = \phi$, we have $\mu(U) \ge \lambda(K) + \lambda(L) \gt \mu(U \cap V) + \mu(U - K) - 2\epsilon \ge \mu(U \cap V) + \mu(U - V) - 2\epsilon$.
Letting $\epsilon \rightarrow 0$, we have $\mu(U) \ge \mu(U\cap V) + \mu(U - V)$ as desired.
Proposition
There exists a unique Radon measure $\mu$ on $X$ such that $\mu = \lambda$ on $\Gamma$.
Proof:
The uniqueness is clear.
The existence follows from Lemma 1 and Lemma 7.
