Wlod AA gave a good counterexample for the case when $K$ is not required to be compact, here I give a counterexample $K$ compact, first in a locally convex space, and then for a(n infinite-dimensional) separable normed space.

There is a standard counterexample if $X$ is only required to be locally convex, which is to take $X = C([0,1])^*$ with the weak-* topology, and to take $K$ to be the set of unital ring homomorphisms $C([0,1]) \rightarrow \mathbb{R}$. Making free use of the Riesz representation theorem to consider elements of $C([0,1])^*$ as measures on $[0,1]$, the elements of $K$ are the Dirac $\delta$-measures. Now, for each element $\mu$ of $\sigma\mbox{-}\mathrm{co}(K)$, there exists a countable set $S \subseteq [0,1]$ such that $\mu([0,1]\setminus S) = 0$. However, $\overline{\mathrm{co}}(K)$ consists of $P([0,1])$, the set of all positive unital linear functionals on $C([0,1])$, *i.e.* all probability measures on $[0,1]$, and so Lebesgue measure is an element of $\overline{\mathrm{co}}(K) \setminus \sigma\mbox{-}\mathrm{co}(K)$. 

To get this to happen in a normed space, we will use $\ell^2$, and embed $P([0,1])$ affinely and continuously into it. First, observe that we can affinely embed $P([0,1])$ into $[0,1]^{\mathbb{N}}$, getting each coordinate by evaluating at $x^n$ (including $n = 0$). This is injective because polynomials are norm dense in $C([0,1])$, and continuous by the definition of the weak-* topology. We can then embed $[0,1]^{\mathbb{N}}$ into $\ell^2$ by the mapping:
$$
f(a)_n = \frac{1}{n+1}a_n
$$
this is affine and continuous from the product topology on $[0,1]^\mathbb{N}$ to the norm topology on $\ell^2$ (in fact, it defines a continuous linear map from the *bounded* weak-* topology on $\ell^\infty$ to the norm topology on $\ell^2$). We use $e$ for the composition of these two embeddings, and it is affine and continuous on $P([0,1])$. 

A continuous injective map from a compact Hausdorff space to a Hausdorff space is a homeomorphism onto its image, and as we also preserved convex combinations by making the embedding affine, we have that $\overline{\mathrm{co}}(e(K)) = e(\overline{\mathrm{co}}(K)) = e(P([0,1]))$, while, taking $\lambda$ to be the element of $P([0,1])$ defined by Lebesgue measure, $e(\lambda) \in e(P([0,1]))$, but $e(\lambda) \not\in e(\sigma\mbox{-}\mathrm{co}(K)) = \sigma\mbox{-}\mathrm{co}(e(K))$.